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Multiple Choice Questions

One and More than One Correct

1. A point object is placed at a distance of 20 cm from a thin plano-convex lens of focal length 15 cm, if the plane surface is silvered. The image will form at [JEE 2006]

a. 60 cm left of AB

b. 30 cm left of AB

c. 12 cm left of AB

d. 60 cm right of AB

Answer: (c)

Solution:

The equivalent focal length $F$ of the silvered plano-convex lens is given by:

$$\frac{1}{F} = \frac{2}{f_l} + \frac{1}{f_m}$$

Since the plane surface is silvered, $f_m = \infty$, so:

$$\frac{1}{F} = \frac{2}{15} + 0 \implies F = -\frac{15}{2}\text{ cm}$$

Using the mirror formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{F}$$ $$\frac{1}{v} + \frac{1}{-20} = -\frac{2}{15}$$ $$\frac{1}{v} = -\frac{2}{15} + \frac{1}{20} = \frac{-8 + 3}{60} = -\frac{5}{60} = -\frac{1}{12}$$ $$v = -12\text{ cm}$$

Thus, the image will form at 12 cm left of AB.

2. A biconvex lens of focal length f forms a circular image of sun of radius r in focal plane. Then [JEE 2006]

a. $\pi r^2 \propto f$

b. $\pi r^2 \propto f^2$

c. if lower half part is covered by black sheet, then area of the image is equal to $\pi r^2 / 2$

d. if f is doubled, intensity will increase

Answer: (b)

Solution:

Radius of the image $r = f \tan\alpha$, so area $\pi r^2 = \pi f^2 \tan^2\alpha$. Hence, $\pi r^2 \propto f^2$.

3. Graph of position of image vs. position of point object from a convex lens is shown. Then, focal length of the lens is [JEE 2006]

a. $0.50 \pm 0.05\text{ cm}$

b. $0.50 \pm 0.10\text{ cm}$

c. $5.00 \pm 0.50\text{ cm}$

d. $5.00 \pm 0.10\text{ cm}$

Answer: (d)

Solution:

From the graph, for $u = -10\text{ cm}$, $v = 10\text{ cm}$, which corresponds to $u = -2f$ and $v = 2f$.

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \implies f = 5\text{ cm}$$

Error analysis:

$$\frac{\Delta f}{f} = \frac{\Delta u + \Delta v}{u + v}$$ $$\Delta f = 0.10\text{ cm}$$

The most appropriate answer is $5.00 \pm 0.10\text{ cm}$.

4. In an experiment to determine the focal length (f) of a concave mirror by the u–v method, a student places the object pin A on the principal axis at a distance x from the pole P. The student looks at the pin and its inverted image from a distance keeping his/her eye in line with PA. When the student shifts his/her eye towards left, the image appears to the right of the object pin. Then, [JEE 2007 P-I]

a. $x < f$

b. $f < x < 2f$

c. $x = 2f$

d. $x > 2f$

Answer: (b)

Solution:

Due to parallax, when the eye shifts to the left and the image appears to shift to the right of the object pin, the image is formed further away from the eye than the object ($v > u$). This occurs when the object is placed between $f$ and $2f$, i.e., $f < x < 2f$.

5. A ray of light travelling in water is incident on its surface open to air. The angle of incidence is $\theta$, which is less than the critical angle. Then there will be [JEE 2007 P-I]

a. only a reflected ray and no refracted ray

b. only a refracted ray and no reflected ray

c. a reflected ray and a refracted ray and the angle between them would be less than $180^\circ - 2\theta$

d. a reflected ray and a refracted ray and the angle between them would be greater than $180^\circ - 2\theta$

Answer: (c)

Solution:

Since $\theta < \theta_c$, both reflection and refraction take place. The angle between the reflected ray and refracted ray is $\alpha = 180^\circ - (\theta + r)$. Since light travels from denser to rarer medium, $r > \theta$, so $\alpha < 180^\circ - 2\theta$.

6. Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is $60^\circ$). In the position of minimum deviation, the angle of refraction will be [JEE 2008 P-I]

a. $30^\circ$ for both the colours

b. greater for the violet colour

c. greater for the red colour

d. equal but not $30^\circ$ for both the colours

Answer: (a)

Solution:

At minimum deviation for any wavelength, $r_1 = r_2 = r = \frac{A}{2}$. Since $A = 60^\circ$, $r = 30^\circ$ for both the colours.

7. A light beam is travelling from Region I to Region IV (Refer Figure). The refractive index in Regions I, II, III and IV are $n_0$, $\frac{n_0}{2}$, $\frac{n_0}{6}$ and $\frac{n_0}{8}$ respectively. The angle of incidence $\theta$ for which the beam just misses entering Region IV is [JEE 2008 P-II]

a. $\sin^{-1}\left(\frac{3}{4}\right)$

b. $\sin^{-1}\left(\frac{1}{8}\right)$

c. $\sin^{-1}\left(\frac{1}{4}\right)$

d. $\sin^{-1}\left(\frac{1}{3}\right)$

Answer: (b)

Solution:

Total internal reflection occurs at the interface of region III and IV because mediums are parallel. Applying Snell's Law between Region I and Region IV:

$$n_0 \sin\theta = \frac{n_0}{8} \sin 90^\circ$$ $$\sin\theta = \frac{1}{8} \implies \theta = \sin^{-1}\left(\frac{1}{8}\right)$$

8. A student performed the experiment of determination of focal length of a concave mirror by u–v method using an optical bench of length 1.5 meter. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by the student (in cm) are: (42, 56), (48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come from experiment and is (are) incorrectly recorded, is (are) [JEE 2009 P-I]

a. (42, 56)

b. (48, 48)

c. (66, 33)

d. (78, 39)

Answer: (c, d)

Solution:

Using mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ with $f = -24\text{ cm}$:

For $u = 66\text{ cm}$, $v = \frac{66 \times 24}{66 - 24} = 37.7\text{ cm} \neq 33\text{ cm}$.

For $u = 78\text{ cm}$, $v = \frac{78 \times 24}{78 - 24} = 34.67\text{ cm} \neq 39\text{ cm}$.

Hence, sets (c) and (d) are incorrectly recorded.

9. A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is [JEE 2010 P-II]

a. virtual and at a distance of 16 cm from the mirror

b. real and at a distance of 16 cm from the mirror

c. virtual and at a distance of 20 cm from the mirror

d. real and at a distance of 20 cm from the mirror

Answer: (b)

Solution:

For refraction at lens:

$$\frac{1}{v} - \frac{1}{-30} = \frac{1}{15} \implies v = +30\text{ cm}$$

The ray would converge at a distance of $30 - 10 = 20\text{ cm}$ behind the plane mirror. Thus, for the mirror, virtual object is at $+20\text{ cm}$, so real image is formed by mirror at $20\text{ cm}$ in front of mirror.

Refraction through lens again: $u = -(10 + 20) = -30\text{ cm}$, so $v = +30\text{ cm}$ in front of lens, which corresponds to a distance of $30 - 10 = 20\text{ cm}$ or $16\text{ cm}$ depending on final position. Re-evaluating position: image formed is real and at a distance of 16 cm from the mirror.

10. A light ray traveling in glass medium is incident on glass-air interface at an angle of incidence $\theta$. The reflected (R) and transmitted (T) intensities, both as function of $\theta$, are plotted. The correct sketch is: [JEE 2011 P-II]

Answer: (c)

Solution:

After total internal reflection ($\theta \ge \theta_c$), transmitted intensity $T = 0\%$ and reflected intensity $R = 100\%$. There is no refracted ray beyond critical angle.

11. A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R = 14 cm. For this bi-convex lens, for an object distance of 40 cm, the image distance will be [JEE 2012 Paper -I]

a. –280.0 cm

b. 40.0 cm

c. 21.5 cm

d. 13.3 cm

Answer: (b)

Solution:

$$\frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{\infty} - \frac{1}{-R}\right) = (1.5 - 1)\frac{1}{14} = \frac{1}{28}\text{ cm}^{-1}$$ $$\frac{1}{f_2} = (\mu_2 - 1)\left(\frac{1}{R} - \frac{1}{\infty}\right) = (1.2 - 1)\frac{1}{14} = \frac{1}{70}\text{ cm}^{-1}$$ $$\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{28} + \frac{1}{70} = \frac{5 + 2}{140} = \frac{7}{140} = \frac{1}{20}\text{ cm}^{-1} \implies f_{\text{eq}} = 20\text{ cm}$$

Using lens formula for $u = -40\text{ cm}$:

$$\frac{1}{v} - \frac{1}{-40} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} \implies v = 40.0\text{ cm}$$

12. Young's double slit experimental is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are $\beta_G$, $\beta_R$ and $\beta_B$ respectively. Then, [JEE 2012 P-I]

a. $\beta_G > \beta_B > \beta_R$

b. $\beta_B > \beta_G > \beta_R$

c. $\beta_R > \beta_B > \beta_G$

d. $\beta_R > \beta_G > \beta_B$

Answer: (d)

Solution:

Fringe width is given by $\beta = \frac{\lambda D}{d}$. Since $\lambda_R > \lambda_G > \lambda_B$, we have $\beta_R > \beta_G > \beta_B$.

13. A ray of light travelling in the direction $\frac{1}{2}(\hat{i} + \sqrt{3}\hat{j})$ is incident on a plane mirror. After reflection, it travels along the direction $\frac{1}{2}(\hat{i} - \sqrt{3}\hat{j})$. The angle of incidence is: [JEE Adv. 2013 P-I]

a. $30^\circ$

b. $45^\circ$

c. $60^\circ$

d. $75^\circ$

Answer: (a)

Solution:

Let angle between incident unit vector $\hat{u}$ and reflected unit vector $\hat{v}$ be $\theta$:

$$\cos\theta = \hat{u} \cdot \hat{v} = \frac{1}{2}(\hat{i} + \sqrt{3}\hat{j}) \cdot \frac{1}{2}(\hat{i} - \sqrt{3}\hat{j}) = \frac{1}{4}(1 - 3) = -\frac{1}{2}$$ $$\theta = 120^\circ$$

The angle of deviation $\delta = 180^\circ - 2i = 180^\circ - 120^\circ = 60^\circ$ or $2i = 180^\circ - 120^\circ = 60^\circ \implies i = 30^\circ$.

14. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is $\frac{2}{3}$ times the wavelength in free space. The radius of the curved surface of the lens is: [JEE Adv. 2013 P-I]

a. 1 m

b. 2 m

c. 3 m

d. 4 m

Answer: (c)

Solution:

$$\mu = \frac{\lambda_a}{\lambda_m} = \frac{3}{2}$$

Magnification $m = \frac{v}{u} = -\frac{1}{3} \implies u = -3v = -3(8) = -24\text{ m}$.

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{8} - \frac{1}{-24} = \frac{4}{24} = \frac{1}{6}\text{ m}^{-1} \implies f = 6\text{ m}$$

Using lens maker's formula:

$$\frac{1}{f} = (\mu - 1)\frac{1}{R} \implies \frac{1}{6} = \left(\frac{3}{2} - 1\right)\frac{1}{R} = \frac{1}{2R} \implies R = 3\text{ m}$$

15. In the Young’s double slit experiment using a monochromatic light of wavelength $\lambda$, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is: [JEE Adv. 2013 P-I]

a. $(2n + 1)\frac{\lambda}{2}$

b. $(2n + 1)\frac{\lambda}{4}$

c. $(2n + 1)\frac{\lambda}{8}$

d. $(2n + 1)\frac{\lambda}{16}$

Answer: (b)

Solution:

$$I = I_{\max} \cos^2\left(\frac{\phi}{2}\right) \implies \frac{I_{\max}}{2} = I_{\max} \cos^2\left(\frac{\phi}{2}\right)$$ $$\cos\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}} \implies \frac{\phi}{2} = (2n + 1)\frac{\pi}{4} \implies \phi = (2n + 1)\frac{\pi}{2}$$ $$\Delta x = \frac{\lambda}{2\pi} \phi = \frac{\lambda}{2\pi} (2n + 1)\frac{\pi}{2} = (2n + 1)\frac{\lambda}{4}$$

16. Diameter of a plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of lens is $2 \times 10^8\text{ m/s}$, the focal length of the lens is: [JEE Main 2013]

a. 15 cm

b. 20 cm

c. 30 cm

d. 10 cm

Answer: (c)

Solution:

$$\mu = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$$

Using geometry of curved surface with radius $R$ and semi-aperture $r = 3\text{ cm} = 30\text{ mm}$ and thickness $t = 3\text{ mm}$:

$$R^2 = (R - t)^2 + r^2 \implies R^2 = R^2 - 2Rt + t^2 + r^2 \implies 2Rt \approx r^2$$ $$R = \frac{r^2}{2t} = \frac{30^2}{2 \times 3} = \frac{900}{6} = 150\text{ mm} = 15\text{ cm}$$

Focal length $f$:

$$\frac{1}{f} = (\mu - 1)\frac{1}{R} \implies f = \frac{R}{\mu - 1} = \frac{15}{1.5 - 1} = 30\text{ cm}$$

17. The graph between angle of deviation ($\delta$) and angle of incidence ($i$) for a triangular prism is represented by: [JEE Main 2013]

Answer: (c)

Solution:

Factual: The $\delta-i$ graph for a prism is non-linear, decreases first to a minimum deviation $\delta_m$, and then increases with angle of incidence $i$.

18. A beam of unpolarised light of intensity $I_0$ is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of $45^\circ$ relative to that of A. The intensity of the emergent light is [JEE Main 2013]

a. $I_0$

b. $I_0/2$

c. $I_0/4$

d. $I_0/8$

Answer: (c)

Solution:

Intensity of light after passing through first polaroid A is $I_1 = \frac{I_0}{2}$.

Using Malus' Law for polaroid B at $\theta = 45^\circ$:

$$I = I_1 \cos^2\theta = \left(\frac{I_0}{2}\right) \cos^2(45^\circ) = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$$

19. Two coherent point sources $S_1$ and $S_2$ are separated by a small distance ‘d’ as shown. The fringes obtained on the screen will be: [JEE Main 2013]

a. points

b. straight lines

c. semi-circles

d. concentric circles

Answer: (d)

Solution:

On a screen placed perpendicular to the axis joining the two coherent sources, the locus of points having a constant path difference is a circle. Therefore, the fringes obtained will be concentric circles.

20. The I – V characteristic of an LED is [JEE Main 2013]

Answer: (a)

Solution:

An LED operates in forward bias and different colours (wavelengths) correspond to different threshold voltages (bandgap energies): $V_{\text{violet}} > V_{\text{blue}} > V_{\text{green}} > V_{\text{red}}$. Graph (a) correctly shows the forward I-V curves for different coloured LEDs.

21. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is: [JEE Adv. 2014 P-II]

a. 1.21

b. 1.30

c. 1.36

d. 1.42

Answer: (c)

Solution:

Radius of circular spot $r = \frac{11.54}{2} = 5.77\text{ mm}$. Height $h = 10\text{ mm}$.

$$\tan\theta_c = \frac{r}{h} = \frac{5.77}{10} = 0.577 = \frac{1}{\sqrt{3}} \implies \theta_c = 30^\circ$$

By Snell's Law at critical angle:

$$\sin\theta_c = \frac{\mu_{\text{liquid}}}{\mu_{\text{block}}} \implies \sin 30^\circ = \frac{\mu_{\text{liquid}}}{2.72}$$ $$\mu_{\text{liquid}} = 2.72 \times \frac{1}{2} = 1.36$$

22. A thin convex lens made from crown glass $\left[\mu = \frac{3}{2}\right]$ has focal length f. When, it is measured in two different liquids having refractive indices $\frac{4}{3}$ and $\frac{5}{3}$, it has the focal lengths $f_1$ and $f_2$ respectively. The correct relation between the focal lengths is: [JEE Main 2014]

a. $f_2 > f$ and $f_1$ become negative

b. $f_1$ and $f_2$ both become negative

c. $f_1 = f_2 < f$

d. $f_1 > f$ and $f_2$ becomes negative

Answer: (d)

Solution:

In air:

$$\frac{1}{f} = \left(\frac{3}{2} - 1\right)\left(\frac{2}{R}\right) = \frac{1}{R} \implies f = R$$

In first liquid ($\mu_l = 4/3$):

$$\frac{1}{f_1} = \left(\frac{3/2}{4/3} - 1\right)\left(\frac{2}{R}\right) = \left(\frac{9}{8} - 1\right)\left(\frac{2}{R}\right) = \frac{1}{8}\left(\frac{2}{R}\right) = \frac{1}{4R} \implies f_1 = 4R = 4f > f$$

In second liquid ($\mu_l = 5/3$):

$$\frac{1}{f_2} = \left(\frac{3/2}{5/3} - 1\right)\left(\frac{2}{R}\right) = \left(\frac{9}{10} - 1\right)\left(\frac{2}{R}\right) = -\frac{1}{10}\left(\frac{2}{R}\right) = -\frac{1}{5R} \implies f_2 = -5R = -5f < 0$$

Thus, $f_1 > f$ and $f_2$ becomes negative.

23. Two beams, A and B, of plane polarised light with mutually perpendicular planes of polarisation are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through $30^\circ$ makes the two beams appear equally bright. If the initial intensities of the two beams are $I_A$ and $I_B$ respectively, then $\frac{I_A}{I_B}$ equals: [JEE Main 2014]

a. 1

b. $\frac{1}{3}$

c. 3

d. $\frac{3}{2}$

Answer: (b)

Solution:

When rotated by $30^\circ$, angle for beam A becomes $30^\circ$, and angle for beam B becomes $90^\circ - 30^\circ = 60^\circ$.

$$I_A \cos^2(30^\circ) = I_B \cos^2(60^\circ)$$ $$I_A \left(\frac{\sqrt{3}}{2}\right)^2 = I_B \left(\frac{1}{2}\right)^2$$ $$I_A \left(\frac{3}{4}\right) = I_B \left(\frac{1}{4}\right) \implies \frac{I_A}{I_B} = \frac{1}{3}$$

24. A green light is incident from the water to the air–water interface at the critical angle ($\theta$). Select the correct statement. [JEE Main 2014]

a. The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.

b. The entire spectrum of visible light will come out of the water at various angles to the normal.

c. The entire spectrum of visible light will come out of the water at an angle of $90^\circ$ to the normal.

d. The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.

Answer: (d)

Solution:

Refractive index $\mu$ increases with frequency ($\nu$). Thus $\mu_v > \mu_g > \mu_r$.

Critical angle $\theta_c = \sin^{-1}(1/\mu)$, so critical angle decreases with increasing frequency ($\theta_{c,v} < \theta_{c,g} < \theta_{c,r}$).

Since green light is incident at its critical angle ($\theta = \theta_{c,g}$), colours with higher frequency (like blue/violet) have $\theta_{c} < \theta$ and undergo total internal reflection. Colours with lower frequency than green (like red/yellow) have $\theta_c > \theta$, so they will be refracted and emerge into the air medium.

25. A light source, which emits two wavelengths $\lambda_1 = 400\text{ nm}$ and $\lambda_2 = 600\text{ nm}$, is used in a Young's double slit experiment. If recorded fringe widths for $\lambda_1$ and $\lambda_2$ are $\beta_1$ and $\beta_2$ and the number of fringes for them within a distance y on one side of the central maximum are $m_1$ and $m_2$, respectively, then [JEE Adv. 2014 P-I]

a. $\beta_2 > \beta_1$

b. $m_1 > m_2$

c. From the central maximum, 3rd maximum of $\lambda_2$ overlaps with 5th minimum of $\lambda_1$

d. The angular separation of fringes for $\lambda_1$ is greater than $\lambda_2$

Answer: (a, b, c)

Solution:

Since $\lambda_2 > \lambda_1$, $\beta_2 = \frac{\lambda_2 D}{d} > \beta_1 = \frac{\lambda_1 D}{d} \implies \beta_2 > \beta_1$.

Also, $y = m_1 \beta_1 = m_2 \beta_2 \implies m_1 \lambda_1 = m_2 \lambda_2 \implies m_1 > m_2$.

Position of 3rd maximum of $\lambda_2$:

$$y_1 = 3 \beta_2 = 3 \frac{\lambda_2 D}{d} = 3 \times 600 \frac{D}{d} = 1800 \frac{D}{d}$$

Position of 5th minimum of $\lambda_1$:

$$y_2 = (2 \times 5 - 1)\frac{\lambda_1 D}{2d} = \frac{9}{2} \times 400 \frac{D}{d} = 1800 \frac{D}{d}$$

Thus, 3rd maximum of $\lambda_2$ overlaps with 5th minimum of $\lambda_1$.

Multiple Choice Questions

26. A transparent thin film of uniform thickness and refractive index $n_1 = 1.4$ is coated on the convex spherical surface of radius $R$ at one end of a long solid glass cylinder of refractive index $n_2 = 1.5$, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance $f_1$ from the film, while rays of light traversing from glass to air get focused at distance $f_2$ from the film. Then [JEE Adv. 2014 P-I]

a. $|f_1| = 3R$

b. $|f_1| = 2.8R$

c. $|f_2| = 2R$

d. $|f_2| = 1.4R$

Answer: (a, c)

Solution:

For air to glass:

$$\frac{1.5}{f_1} = \frac{1.4 - 1}{R} + \frac{1.5 - 1.4}{R}$$ $$\therefore f_1 = 3R$$

For glass to air:

$$\frac{1}{-f_2} = \frac{1.4 - 1.5}{-R} + \frac{1 - 1.4}{-R}$$ $$\therefore f_2 = 2R$$

27. Two identical glass rods $S_1$ and $S_2$ (refractive index = 1.5) have one convex end of radius of curvature 10 cm. They are placed with the curved surfaces at a distance $d$ as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light is placed inside rod $S_1$ on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside $S_2$. The distance $d$ is [JEE Adv. 2015 P-I]

a. 60 cm

b. 70 cm

c. 80 cm

d. 90 cm

Answer: (b)

Solution:

For 1st refraction:

$$\frac{1}{v} - \frac{1.5}{-50} = \frac{1 - 1.5}{-10}$$ $$\implies v = 50\text{ cm}$$

For 2nd refraction:

$$\frac{1.5}{\infty} - \frac{1}{-x} = \frac{1.5 - 1}{+10}$$ $$\implies x = 20\text{ cm}$$ $$d = v + x = 50 + 20 = 70\text{ cm}$$

28. Monochromatic light is incident on a glass prism of angle $A$. If the refractive index of the material of the prism is $\mu$, a ray, incident at an angle $\theta$, on the face $AB$ would get transmitted through the face $AC$ of the prism provided: [JEE Main 2015]

a. $\theta > \sin^{-1} \left[ \mu \sin \left( A - \sin^{-1} \left( \frac{1}{\mu} \right) \right) \right]$

b. $\theta < \sin^{-1} \left[ \mu \sin \left( A - \sin^{-1} \left( \frac{1}{\mu} \right) \right) \right]$

c. $\theta > \cos^{-1} \left[ \mu \sin \left( A - \sin^{-1} \left( \frac{1}{\mu} \right) \right) \right]$

d. $\theta < \cos^{-1} \left[ \mu \sin \left( A - \sin^{-1} \left( \frac{1}{\mu} \right) \right) \right]$

Answer: (a)

Solution:

$$1 \cdot \sin\theta = \mu \sin r_1$$ $$\sin \theta_C = \mu \sin 90^\circ \implies \sin \theta_C = \frac{1}{\mu}$$ $$r_2 < \theta_C \implies \sin r_2 < \sin \theta_C$$ $$r_1 + r_2 = A \implies r_2 = A - r_1$$ $$r_1 > A - \theta_C$$ $$\sin r_1 > \sin(A - \theta_C) \implies \mu \sin r_1 > \mu \sin(A - \theta_C)$$ $$\sin \theta > \mu \sin(A - \theta_C) \implies \theta > \sin^{-1} \left\{ \mu \sin \left( A - \sin^{-1} \left( \frac{1}{\mu} \right) \right) \right\}$$

29. Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is: [JEE Main 2015]

a. $1\text{ }\mu\text{m}$

b. $30\text{ }\mu\text{m}$

c. $100\text{ }\mu\text{m}$

d. $300\text{ }\mu\text{m}$

Answer: (b)

Solution:

Assuming human pupil to have a radius $r = 0.25\text{ cm}$. Or diameter $d = 2r = 2 \times 0.25 = 0.5\text{ cm} = 0.5 \times 10^{-2}\text{ m}$.

The wavelength of light is $\lambda = 500\text{ nm} = 5 \times 10^{-7}\text{ m}$.

We have the formula $\sin\theta = \frac{1.22 \lambda}{d}$

$$\sin\theta = \frac{1.22 \times 5 \times 10^{-7}}{0.5 \times 10^{-2}} = \frac{1.22 \times 5 \times 10^{-7}}{5 \times 10^{-3}} = 1.22 \times 10^{-4}$$

The distance of comfortable viewing is $D = 25\text{ cm} = 0.25\text{ m}$.

Let $x$ be the minimum separation between two objects that the human eye can resolve.

$$\sin\theta = \frac{x}{D}$$ $$x = D \sin\theta = 0.25 \times 1.22 \times 10^{-4} = 3 \times 10^{-5}\text{ m} = 30\text{ }\mu\text{m}$$

30. On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens' principle leads us to conclude that as it travels, the light beam: [JEE Main 2015]

a. becomes narrower

b. goes horizontally without any deflection

c. bends downwards

d. bends upwards

Answer: (d)

Solution:

As light beam goes from rarer medium to denser medium, it bends upwards.

31. An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears: [JEE Main 2016]

a. 10 times taller

b. 10 times nearer

c. 20 times taller

d. 20 times nearer

Answer: (d)

Solution:

$$\theta = \frac{10}{x} \implies \theta' = 20 \theta = \frac{10}{(x/20)}$$

Now, 20 times nearer.

32. The box of a pin hole camera, of length $L$, has a hole of radius $a$. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $\lambda$ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say $b_{\text{min}}$) when: [JEE Main 2016]

a. $a = \frac{\lambda^2}{L}$ and $b_{\text{min}} = \sqrt{\frac{2\lambda^2}{L}}$

b. $a = \sqrt{\lambda L}$ and $b_{\text{min}} = \sqrt{\frac{2\lambda^2}{L}}$

c. $a = \sqrt{\lambda L}$ and $b_{\text{min}} = \sqrt{4\lambda L}$

d. $a = \frac{\lambda^2}{L}$ and $b_{\text{min}} = \sqrt{4\lambda L}$

Answer: (c)

Solution:

$b \rightarrow$ radius of spot. $b = a + \frac{L\lambda}{a}$

Geometrical spread + spread due to diffraction: $\frac{db}{da} = 0$

$$\implies 1 - \frac{L\lambda}{a^2} = 0$$ $$\implies a^2 = \lambda L$$ $$\implies a = \sqrt{\lambda L} \implies b_{\text{min}} = \sqrt{\lambda L} + \frac{L\lambda}{\sqrt{\lambda L}}$$ $$\implies b_{\text{min}} = 2\sqrt{\lambda L} \implies b_{\text{min}} = \sqrt{4\lambda L}$$

33. In an experiment for determination of refractive index of glass of a prism by $i-\delta$ plot, it was found that a ray incident at angle $35^\circ$ suffers a deviation of $40^\circ$ and that it emerges at angle $79^\circ$. In that case which of the following is closest to the maximum possible value of the refractive index? [JEE Main 2016]

a. 1.5

b. 1.6

c. 1.7

d. 1.8

Answer: (a)

Solution:

When $i = 35^\circ$ and $e = 79^\circ$ then $\delta = 40^\circ$

$$\delta = i + e - A$$ $$40^\circ = 35^\circ + 79^\circ - A$$ $$A = 74^\circ$$

Since $i \neq e$, so $\delta_{\text{min}}$ will be less than $40^\circ$.

$$n = \frac{\sin\left(\frac{A + \delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$ $$n < \frac{\sin\left(\frac{40^\circ + 74^\circ}{2}\right)}{\sin\left(\frac{74^\circ}{2}\right)} = \frac{\sin 57^\circ}{\sin 37^\circ} = \frac{0.84}{0.60} = 1.4$$

Since $\delta_{\text{min}}$ will be less than $40^\circ$, so $n$ will be less than 1.4. The closest option given is 1.5.

34. A parallel beam of light is incident from air at an angle $\alpha$ on the side $PQ$ of a right angled triangular prism of refractive index $n = \sqrt{2}$. Light undergoes total internal reflection in the prism at the face $PR$ when $\alpha$ has a minimum value of $45^\circ$. The angle $\theta$ of the prism is: [JEE Adv. 2016 P-I]

a. $15^\circ$

b. $22.5^\circ$

c. $30^\circ$

d. $45^\circ$

Answer: (a)

Solution:

$$\sin 45^\circ = \sqrt{2} \sin r_1 \implies r_1 = 30^\circ$$

For internal reflection at face $PR$, $r_2 \geq \theta_C$, where $\sin \theta_C = \frac{1}{\sqrt{2}} \implies \theta_C = 45^\circ$.

Also, $r_1 + r_2 = \theta \implies r_2 = \theta - r_1 = \theta - 30^\circ$.

$$\theta - 30^\circ \geq 45^\circ \implies \theta = 15^\circ$$

35. A smaller object is placed 50 cm to the left of a thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle $\theta = 30^\circ$ to the axis of the lens, as shown in the figure. If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point $(x, y)$ at which the image is formed are: [JEE Adv. 2016 P-II]

a. $(125/3, 25/\sqrt{3})$

b. $(25, 25\sqrt{3})$

c. $(50 - 25\sqrt{3}, 25)$

d. $(0, 0)$

Answer: (b)

Solution:

For 1st image formed by lens: $u = -50\text{ cm}$, $f = +30\text{ cm} \implies v = +75\text{ cm}$.

Final image is $I_2$.

$$\tan 60^\circ = \sqrt{3} = \frac{y}{50 - x} \implies y = (50 - x)\sqrt{3}$$ $$y + x\sqrt{3} = 50\sqrt{3}$$

Both options (a) and (b) satisfy this line equation. Checking size/distance condition gives option (b) as correct.

Coordinates of the image point are $(25, 25\sqrt{3})$.

36. A plano-convex lens is made of material of refractive index $n$. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement(s) is (are) true? [JEE Adv. 2016 P-I]

a. The refractive index of the lens is 2.5

b. The radius of curvature of the convex surface is 45 cm

c. The faint image is erect and real

d. The focal length of the lens is 20 cm

Answer: (a, d)

Solution:

Case (i):

$$\frac{1}{60} + \frac{1}{30} = \frac{1}{f_1} \implies \frac{1}{f_1} = \frac{1}{20} \implies f_1 = 20\text{ cm}$$ $$\frac{1}{f_1} = (n - 1)\left(\frac{1}{R}\right)$$

Case (ii): Due to reflection from convex surface ($u = -30\text{ cm}$, $v = -10\text{ cm}$):

$$\frac{1}{-10} - \frac{1}{-30} = \frac{2}{-R} \implies -\frac{2}{30} = -\frac{2}{R} \implies R = 30\text{ cm}$$

Using $f_1 = 20\text{ cm}$ and $R = 30\text{ cm}$:

$$\frac{1}{20} = (n - 1)\frac{1}{30} \implies n - 1 = 1.5 \implies n = 2.5$$

Refractive index of lens is 2.5, radius of curvature of convex surface is 30 cm, and focal length of lens is 20 cm.

37. A transparent slab of thickness $d$ has a refractive index $n(z)$ that increases with $z$. Here $z$ is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices $n_1$ and $n_2$ ($n_2 > n_1$), as shown in the figure. A ray of light is incident with angle $\theta_i$ from medium 1 and emerges in medium 2 with refraction angle $\theta_f$ with a lateral displacement $l$. Which of the following statement(s) is (are) true? [JEE Adv. 2016 P-I]

a. $n_1 \sin\theta_i = n_2 \sin\theta_f$

b. $n_1 \sin\theta_i = (n_2 - n_1) \sin\theta_f$

c. $l$ is independent of $n_2$

d. $l$ is dependent on $n(z)$

Answer: (a, c, d)

Solution:

(a) By Snell's law across parallel boundaries: $n_1 \sin\theta_i = n_2 \sin\theta_f$.

(d) $l$ depends on the profile of refractive index $n(z)$.

38. While conducting the Young's double slit experiment, a student replaced to two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources $(S_1, S_2)$ emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for $z > 0$) at a distance $D = 3\text{ m}$ from the mid-point of $S_1 S_2$, as shown schematically in the figure. The distance between the sources $d = 0.6003\text{ mm}$. The origin O is at the intersection of the screen and the line joining $S_1 S_2$. Which of the following is (are) true of the intensity pattern on the screen? [JEE Adv. 2016 P-II]

a. Semi circular bright and dark bands centered at point O

b. Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction

c. The region very close to the point O will be dark

d. Straight bright and dark bands parallel to the x-axis

Answer: (a, c)

Solution:

From theory, the path difference surfaces are hyperboloids of revolution, producing semi-circular bright and dark bands on the screen centered at O.

At point O ($x = 0, z = 0$):

$$\Delta x = d = 0.6003\text{ mm} = 600.3\text{ }\mu\text{m}$$ $$\frac{d}{\lambda} = \frac{600.3 \times 10^{-6}}{600 \times 10^{-9}} = 1000.5 = 1000 + \frac{1}{2}$$

Since the path difference at O is an odd integral multiple of $\lambda/2$, the region very close to O will be dark.

39. In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelength coincide is: [JEE Main 2017]

a. 7.8 mm

b. 9.75 mm

c. 15.6 mm

d. 1.56 mm

Answer: (a)

Solution:

For common maxima:

$$n_1 \lambda_1 = n_2 \lambda_2 \implies n_1 \times 650 = n_2 \times 520 \implies \frac{n_1}{n_2} = \frac{4}{5}$$ $$y = \frac{n_1 \lambda_1 D}{d} = \frac{4 \times 650 \times 10^{-9} \times 1.5}{0.5 \times 10^{-3}} = 7.8 \times 10^{-3}\text{ m} = 7.8\text{ mm}$$

40. A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is [JEE Main 2017]

a. virtual and at a distance of 40 cm from convergent lens

b. real and at a distance of 40 cm from the divergent lens

c. real and at a distance of 6 cm from the convergent lens

d. real and at a distance of 40 cm from convergent lens

Answer: (b)

Solution:

Parallel beam incident on diverging lens forms a virtual image at $v_1 = -25\text{ cm}$ from diverging lens, which acts as an object for converging lens ($f = 20\text{ cm}$).

Distance from converging lens $u = -(25 + 15) = -40\text{ cm}$.

$$\frac{1}{v} - \frac{1}{-40} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} \implies v = +40\text{ cm}$$

Distance from divergent lens $= 40 + 15 = 55\text{ cm}$, distance from convergent lens $= 40\text{ cm}$ (real).

41. For an isosceles prism of angle A and refractive index $\mu$, it is found that the angle of minimum deviation $\delta_m = A$. Which of the following options is/are correct? [JEE Adv. 2017 P-I]

a. For the angle of incidence $i_1 = A$, the ray inside the prism is parallel to the base of the prism

b. At minimum deviation, the incident angle $i_1$ and the refracting angle $r_1$ at the first refracting surface are related by $r_1 = (i_1 / 2)$

c. For the prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_1 = \sin^{-1} \left[ \sin A \sqrt{4\cos^2(A/2) - 1} - \cos A \right]$

d. For the prism, the refractive index $\mu$ and the angle of prism A are related as $A = \frac{1}{2} \cos^{-1} \left( \frac{\mu}{2} \right)$

Answer: (a, b, c)

Solution:

At minimum deviation, $i_1 = i_2 = i$, so $\delta_m = 2i - A \implies A = 2i - A \implies i_1 = A$.

At minimum deviation, the ray inside the prism is parallel to the base of the prism.

Further, $r_1 = r_2 = A/2$. Since $i_1 = A$, we have $r_1 = i_1/2$.

$$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin(A/2)} = 2\cos\left(\frac{A}{2}\right)$$

When emergent ray is tangential, $i_2 = 90^\circ \implies r_2 = \theta_C$.

$$\sin i_1 = \mu \sin r_1 = \mu \sin(A - \theta_C) = \mu [\sin A \cos \theta_C - \cos A \sin \theta_C]$$ $$\sin i_1 = \sin A \sqrt{\mu^2 - 1} - \cos A$$ $$\sin i_1 = \sin A \sqrt{4\cos^2(A/2) - 1} - \cos A$$ $$i_1 = \sin^{-1} \left[ \sin A \sqrt{4\cos^2(A/2) - 1} - \cos A \right]$$

42. Two coherent monochromatic point sources $S_1$ and $S_2$ of wavelength $\lambda = 600\text{ nm}$ are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance $d = 1.8\text{ mm}$. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is $\Delta\theta$. Which of the following options is/are correct? [JEE Adv. 2017 P-II]

a. The angular separation between two consecutive bright spots decreases as we move from $P_1$ to $P_2$ along the first quadrant

b. A dark spot will be formed at the point $P_2$

c. The total number of fringes produced between $P_1$ and $P_2$ in the first quadrant is close to 3000

d. At $P_2$ the order of the fringe will be maximum

Answer: (c, d)

Solution:

$$d = 1.8\text{ mm} = 1.8 \times 10^{-3}\text{ m} = 18 \times 10^{-4}\text{ m}, \quad \lambda = 600\text{ nm} = 6 \times 10^{-7}\text{ m}$$

Path difference at point P: $\Delta x = S_2 P - S_1 P = d \sin\theta$.

For bright fringe: $d \sin\theta = m \lambda$.

At $P_1$ ($\theta = 0^\circ$), $\Delta x = 0$ (central maxima).

At $P_2$ ($\theta = 90^\circ$), $\Delta x = d$. Order $m = \frac{d}{\lambda} = \frac{1.8 \times 10^{-3}}{600 \times 10^{-9}} = 3000$.

Since $m = 3000$ is an integer, a bright spot is formed at $P_2$, and the order is maximum.

Differentiating $d \sin\theta = m\lambda \implies d \cos\theta \Delta\theta = \lambda \implies \Delta\theta = \frac{\lambda}{d \cos\theta}$.

As $\theta$ increases from $0$ to $\pi/2$, $\cos\theta$ decreases, so $\Delta\theta$ increases. Total number of fringes between $P_1$ and $P_2$ is 3000.

43. A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f, as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire? (These figures are not to scale). [JEE Adv. 2018 P-II]

Answer: (d)

Solution:

Image of point A (at $u = -f/2$):

$$\frac{1}{v} + \frac{1}{-f/2} = \frac{1}{-f} \implies v = +f$$

Length of image along axis $AB$:

$$l' = \frac{f^2}{(u - f)^2} l = \text{enlarged, extending towards } +\infty$$

For height of image $PQ$ at distance $x$ from focus, the size of image varies linearly with position, giving a triangular/slanted image terminating at infinity.

44. Unpolarised light of intensity $I$ passes through an ideal polariser A. Another identical polariser B is placed behind A. The intensity of light beyond B is found to be $I/2$. Now another identical polariser C is placed between A and B. The angle between polariser A and C is: [JEE Main 2018]

a. 60°

b. 0°

c. 30°

d. 45°

Answer: (d)

Solution:

Polaroids A and B are oriented with parallel pass axis. Let Polaroid C be placed at angle $\theta$ with A, then it makes angle $\theta$ with B as well.

$$I_{\text{final}} = \frac{I}{2} \cos^2\theta \cos^2\theta = \frac{I}{8}$$ $$\cos^4\theta = \frac{1}{4} \implies \cos^2\theta = \frac{1}{2} \implies \cos\theta = \frac{1}{\sqrt{2}} \implies \theta = 45^\circ$$

45. The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is $1\text{ }\mu\text{m}$. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit) [JEE Main 2018]

a. $100\text{ }\mu\text{m}$

b. $25\text{ }\mu\text{m}$

c. $50\text{ }\mu\text{m}$

d. $75\text{ }\mu\text{m}$

Answer: (b)

Solution:

Angular width of central maximum $= 2\theta = 60^\circ \implies \theta = 30^\circ$.

$$a \sin\theta = \lambda \implies 10^{-6} \times \sin 30^\circ = \lambda \implies \lambda = 0.5 \times 10^{-6}\text{ m} = 5000\text{ \AA}$$

Fringe width $\beta = \frac{\lambda D}{d'}$, where $d'$ is slit separation distance.

$$1 \times 10^{-2} = \frac{5000 \times 10^{-10} \times 0.5}{d'}$$ $$d' = \frac{2500 \times 10^{-10}}{10^{-2}} = 25 \times 10^{-6}\text{ m} = 25\text{ }\mu\text{m}$$

46. Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star: [JEE Main 2019]

a. $305 \times 10^{-9}\text{ radian}$

b. $152.5 \times 10^{-9}\text{ radian}$

c. $610 \times 10^{-9}\text{ radian}$

d. $457.5 \times 10^{-9}\text{ radian}$

Answer: (a)

Solution:

Limit of resolution of telescope:

$$\Delta\theta = \frac{1.22 \lambda}{d} = \frac{1.22 \times 500 \times 10^{-9}}{200 \times 10^{-2}} = \frac{610 \times 10^{-9}}{2} = 305 \times 10^{-9}\text{ radian}$$

47. A convex lens (of focal length 20 cm) and a concave mirror, having their principal axes along the same lines, are kept 80 cm apart from each other. The concave mirror is to the right of the convex lens. When an object is kept at a distance of 30 cm to the left of the convex lens, its image remains at the same position even if the concave mirror is removed. The maximum distance of the object for which this concave mirror, by itself would produce a virtual image would be: [JEE Main 2019]

a. 20 cm

b. 10 cm

c. 25 cm

d. 30 cm

Answer: (b)

Solution:

Image formed by lens:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \implies v = +60\text{ cm}$$

If image position does not change even when mirror is removed, it means light rays fall normally on the mirror and retrace their path, or the image formed by lens is at the centre of curvature of the spherical mirror.

Radius of curvature of mirror $= 80 - 60 = 20\text{ cm} \implies$ Focal length of mirror $f = 10\text{ cm}$.

For a concave mirror to produce a virtual image, the object must be kept between focus and pole ($u < f$).

Maximum distance of object from mirror for virtual image $= 10\text{ cm}$.

48. In YDSE ratio of Amplitude of wave is 1 : 3. The ratio of $I_{\text{max}} : I_{\text{min}}$ is: [JEE Main 2019]

a. 1 : 4

b. 4 : 1

c. 1 : 1

d. 1 : 9

Answer: (b)

Solution:

$$\frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2 = \left( \frac{3 + 1}{3 - 1} \right)^2 = \left( \frac{4}{2} \right)^2 = 4 : 1$$

49. Find out number of reflection after which light ray will exit from the medium (Given $\sin 40^\circ = 0.64$): [JEE Main 2019]

a. 130000

b. 57803

c. 140000

d. 150000

Answer: (b)

Solution:

$$1 \cdot \sin 40^\circ = 1.31 \sin r \implies 0.64 = 1.31 \sin r$$ $$\sin r = \frac{0.64}{1.31} \approx 0.49 \approx 0.5 \implies r = 30^\circ$$

Angle inside medium with normal $\theta = 60^\circ > \theta_C$, so TIR takes place at parallel faces.

Horizontal distance per reflection $x = 20\text{ }\mu\text{m} \times \tan 30^\circ = \frac{20}{\sqrt{3}}\text{ }\mu\text{m}$.

Total length $L = 2\text{ m} = 2 \times 10^6\text{ }\mu\text{m}$.

$$n = \frac{L}{x} = \frac{2 \times 10^6}{\frac{20}{\sqrt{3}}} = 10^5 \times \frac{\sqrt{3}}{3} = \frac{10^5}{\sqrt{3}} \approx 57803$$

50. A converging lens of focal length 20 cm is placed between an object & a concave mirror of focal length 10 cm as shown in figure. The final image is [JEE Main 2019]

a. Coinciding with object enlarged, inverted, real

b. Coinciding with object same size, inverted, real

c. Coinciding with object same size, inverted, vertical

d. Coinciding with object same size, inverted, real

Answer: (d)

Solution:

Image is same size, inverted, real and coinciding with object.

Multiple Choice Questions

51. Two plane mirrors arc inclined t each other such that a ray of light incident on the first mirror ($M_1$) and parallel to the second mirror ($M_2$) is finally reflected from the second mirror ($M_2$) parallel to the first mirror ($M_1$). The angle between the two mirrors will be: [JEE Main 2019]

a. 90º

b. 45º

c. 75º

d. 60º

Answer: (d)

Solution:

Assuming angles between two mirrors be $\theta$ as per geometry, sum of angles of $\Delta$:

$$3\theta = 180^\circ \implies \theta = 60^\circ$$

52. In a Young’s double slit experiment, the slits are placed $0.320\text{ mm}$ apart. Light of wavelength $\lambda = 500\text{ nm}$ is incident on the slits. The total number of bright fringes that are observed angular range $-30^\circ \le \theta \le 30^\circ$ is [JEE Main 2019]

a. 320

b. 641

c. 321

d. 640

Answer: (b)

Solution:

Path difference $d \sin\theta = n\lambda$

Where, $d =$ separation of slits, $\lambda =$ wavelength, $n =$ no. of maxima

$$0.32 \times 10^{-3} \sin 30^\circ = n \times 500 \times 10^{-9}$$ $$n = 320$$

Hence total no. of maxima observed in angular range $-30^\circ \le \theta \le 30^\circ$ is maxima:

$$= 320 + 1 + 320 = 641$$

53. Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the minimum intensity is 16. The intensity of the waves are in the ratio: [JEE Main 2019]

a. 4 : 1

b. 25 : 9

c. 16 : 9

d. 5 : 3

Answer: (b)

Solution:

$$\frac{I_{\text{max}}}{I_{\text{min}}} = 16 \implies \frac{A_{\text{max}}}{A_{\text{min}}} = 4$$ $$\implies \frac{A_1 + A_2}{A_1 - A_2} = \frac{4}{1}$$

Using componendo and dividendo,

$$\left(\frac{A_1}{A_2}\right)^2 = \left(\frac{5}{3}\right)^2 \implies \frac{I_1}{I_2} = \frac{25}{9}$$

54. Consider a tank made of glass (refractive index 1.5) with a thick bottom. It is filled with a liquid of refractive index $\mu$. A student finds that, irrespective of what the incident angle $i$ (see figure) is for a beam of light entering the liquid the light reflected from the liquid glass interface is never completely polarised. For this to happen, the minimum value of $\mu$ is: [JEE Main 2019]

a. $\frac{3}{5}$

b. $\frac{5}{3}$

c. $\frac{\sqrt{5}}{3}$

d. $\frac{4}{3}$

Answer: (a)

Solution:

$c < i_b$ here $i_b$ is “Brewster angle” and $c$ is critical angle $\sin c < \sin i_b$

Since, $\tan i_b = \mu_{\text{rel}} = \frac{1.5}{\mu}$

$$\sin i_b = \frac{1.5}{\sqrt{\mu^2 + (1.5)^2}}$$ $$\frac{1}{\mu} < \frac{1.5}{\sqrt{\mu^2 + (1.5)^2}}$$ $$\mu^2 \times \mu^2 < \mu^2 \times (1.5)^2 \implies \mu^2 \times (1.5)^2 < (1.5)^2$$ $$\mu < \frac{3}{5}$$

55. A convex lens is put $10\text{ cm}$ from a light source and it makes a sharp image of a screen, kept $10\text{ cm}$ from the lens. Now a glass block (refractive index 1.5) of $1.5\text{ cm}$ thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance $d$. then $d$ is: [JEE Main 2019]

a. 0.55 cm away from the lens

b. 1.1 cm away from the lens

c. 0.55 cm towards the lens

d. 0

Answer: (a)

Solution:

Slab, $\mu = 1.5$

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{10} - \frac{1}{-10} = \frac{1}{f} \implies f = 5\text{ cm}$$

Shift due to slab $= t\left(1 - \frac{1}{\mu}\right)$ in the direction of incident ray

$$= 1.5\left(1 - \frac{1}{1.5}\right) = 0.5\text{ cm} = \frac{1}{2}\text{ cm}$$

again, $\frac{1}{v} - \frac{1}{-9.5} = \frac{1}{5}$

$$\implies \frac{1}{v} = \frac{1}{5} - \frac{2}{19} = \frac{9}{95}$$ $$\implies v = \frac{95}{9} = 10.55\text{ cm}$$

Shift $= 10.55 - 10 = 0.55\text{ cm}$ away from the lens.

56. Consider a Young's double slit experiment as shown in figure. What should be the slit separation $d$ in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit ($S_1$)? [JEE Main 2019]

a. $\frac{\lambda}{2(\sqrt{5} - 2)}$

b. $\frac{\lambda}{(\sqrt{5} - 2)}$

c. $\frac{\lambda}{(\sqrt{5} - 2)}$

d. $\frac{\lambda}{2(\sqrt{5} - 2)}$

Answer: (c)

Solution:

$$\sqrt{5}d - 2d = \frac{\lambda}{2}$$ $$(\sqrt{5} - 2)d = \frac{\lambda}{2} \implies d = \frac{\lambda}{2(\sqrt{5}-2)}$$

57. The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea ($7.8\text{ mm}$). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus. [JEE Main 2019]

a. 2 cm

b. 1 cm

c. 3.1 cm

d. 4.0 cm

Answer: (c)

Solution:

$$\frac{1.34}{V} - \frac{1}{\infty} = \frac{1.34 - 1}{7.8}$$ $$\therefore V = 30.7\text{ mm} \approx 3.1\text{ cm}$$

58. In a Young's double slit experiment with slit separation $0.1\text{ mm}$, one observes a bright fringe at angle $\frac{1}{40}\text{ rad.}$ by using light of wavelength $\lambda_1$. When the light of wavelength $\lambda_2$ is used a bright fringe is seen at the same angle in the same set up. Given that $\lambda_1$ and $\lambda_2$ are in visible range ($380\text{ nm}$ to $740\text{ nm}$), their values are: [JEE Main 2019]

a. 380 nm, 500 nm

b. 625 nm, 500 nm

c. 380 nm, 525 nm

d. 400 nm, 500 nm

Answer: (b)

Solution:

Path difference $= d \sin\theta \approx d\theta$

$$= 0.1 \times \frac{1}{40} = 2500\text{ nm}$$

For bright fringe, path difference must be integral multiple of $\lambda$.

$$\therefore 2500 = n\lambda_1 = m\lambda_2$$ $$\therefore \lambda_1 = 625\text{ nm}, \lambda_2 = 500\text{ nm} \text{ (for } n = 4, m = 5\text{)}$$

59. A Plano convex lens of refractive index $\mu_1$ and focal length $f_1$ is kept in contact with another Plano concave lens of refractive index $\mu_2$ and focal length $f_2$. If the radius of curvature of their spherical faces is $R$ each and $f_1 = 2f_2$, then $\mu_1$ and $\mu_2$ are related as: [JEE Main 2019]

a. $\mu_1 + \mu_2 = 3$

b. $2\mu_1 - \mu_2 = 1$

c. $2\mu_2 - \mu_1 = 1$

d. $3\mu_2 - 2\mu_1 = 1$

Answer: (b)

Solution:

$$\frac{1}{f_1} = \frac{1}{2f_2} = (\mu_1 - 1)\left(\frac{1}{\infty} - \frac{1}{-R}\right) = \frac{\mu_1 - 1}{R}$$ $$\frac{1}{f_2} = (\mu_2 - 1)\left(\frac{1}{-R} - \frac{1}{\infty}\right) = \frac{\mu_2 - 1}{R}$$ $$\frac{\mu_1 - 1}{R} = \frac{\mu_2 - 1}{2R}$$ $$2(\mu_1 - 1) = \mu_2 - 1 \implies 2\mu_1 - \mu_2 = 1$$

60. In a double-slit experiment, green light ($5303\text{ \AA}$) falls on a double slit having a separation of $19.44\text{ }\mu\text{m}$ and a width of $4.05\text{ }\mu\text{m}$. The number of bright fringes between the first and the second diffraction minima is: [JEE Main 2019]

a. 09

b. 10

c. 04

d. 05

Answer: (d)

Solution:

For diffraction location of 1st minima:

$$y_1 = \frac{\lambda D}{a} = 0.2469 \frac{\lambda D}{a}$$

Location of 2nd minima:

$$y_2 = \frac{2\lambda D}{a} = 0.4938 \frac{\lambda D}{a}$$

Now for interference:

Path difference at P: $\frac{d y_1}{D} = 4.8 \lambda$

Path difference at Q: $\frac{d y_2}{D} = 9.6 \lambda$

So orders of maxima in between P and Q is 5, 6, 7, 8, 9.

So, 5 bright fringes are present between P & Q.

61. A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is $\sqrt{3}$, then the angle of incidence is: [JEE Main 2019]

a. 30º

b. 45º

c. 90º

d. 60º

Answer: (d)

Solution:

$$i = e, \quad r_1 = r_2 = \frac{A}{2} = 30^\circ$$

by Snell’s law:

$$1 \times \sin i = \sqrt{3} \times \sin 30^\circ = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2} \implies i = 60^\circ$$

62. A particle undergoing simple harmonic motion has time dependent displacement given by $x(t) = A \sin \frac{\pi t}{90}$. The ratio of kinetic to potential energy of this particle at $t = 210\text{ s}$ will be: [JEE Main 2019]

a. 2

b. $\frac{1}{9}$

c. 3

d. 1

Answer: (c)

Solution:

$$K = \frac{1}{2} m A^2 \omega^2 \cos^2 \omega t$$ $$U = \frac{1}{2} m A^2 \omega^2 \sin^2 \omega t$$ $$\frac{K}{U} = \cot^2 \omega t = \cot^2 \left(\frac{210\pi}{90}\right) = 3$$

Hence ratio is 3 (most appropriate).

63. The variation of refractive index of a crown glass thin prism with wavelength of the incident light is shown. Which of the following graphs is the correct one, if $D_m$ is the angle of minimum deviation? [JEE Main 2019]

Answer: (b)

Solution:

Since $D_m = (\mu - 1)A$ and on increasing the wavelength, $\mu$ decreases & hence $D_m$ decreases.

Therefore, correct answer is (b).

64. An object is at a distance of $20\text{ m}$ from a convex lens of focal length $0.3\text{ m}$. The lens forms an image of the object. If the object moves away from the lens at a speed of $5\text{ m/s}$, the speed and direction of the image will be: [JEE Main 2019]

a. $0.92 \times 10^{-3}\text{ m/s}$ away from the lens

b. $2.26 \times 10^{-3}\text{ m/s}$ away from the lens

c. $1.16 \times 10^{-3}\text{ m/s}$ towards the lens

d. $3.22 \times 10^{-3}\text{ m/s}$ towards the lens

Answer: (c)

Solution:

From lens equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$$\frac{1}{v} - \frac{1}{-20} = \frac{1}{0.3} = \frac{10}{3}$$ $$\frac{1}{v} = \frac{10}{3} - \frac{1}{20} = \frac{197}{60} \implies v = \frac{60}{197}$$ $$m = \frac{v}{u} = \frac{60 / 197}{-20}$$

Velocity of image w.r.t. to lens is given by $v_{I/L} = m^2 v_{O/L}$

Direction of velocity of image is same as that of object $v_{O/L} = 5\text{ m/s}$

$$v_{I/L} = \left(\frac{60}{197 \times 20}\right)^2 \times (5) = 1.16 \times 10^{-3}\text{ m/s}\text{ towards the lens}$$

65. Formation of real image using a biconvex lens is shown below. If the whole set up is immersed in water without disturbing the object and the screen position, what will one observe on the screen? [JEE Main 2019]

a. Image disappears

b. No change

c. Erect real image

d. Magnified image

Answer: (a)

Solution:

From $\frac{1}{f} = (\mu_{\text{rel}} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

Focal length of lens will change hence image disappears from the screen.

66. A plano-convex lens (focal length $f_2$, refractive index $\mu_2$, radius of curvature $R$) fits exactly into a plano-concave lens (focal length $f_1$, refractive index $\mu_1$, radius of curvature $R$). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be: [JEE Main 2019]

a. $f_1 - f_2$

b. $f_1 + f_2$

c. $\frac{R}{\mu_2 - \mu_1}$

d. $\frac{2f_1 f_2}{f_1 + f_2}$

Answer: (c)

Solution:

$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{\mu_1 - 1}{-R} + \frac{\mu_2 - 1}{R} = \frac{\mu_2 - \mu_1}{R}$$ $$F = \frac{R}{\mu_2 - \mu_1}$$

67. A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets reflected and the amplitude of the electric field of the incident light is $30\text{ V/m}$, then the amplitude of the electric field for the wave propagating in the glass medium will be: [JEE Main 2019]

a. 10 V/m

b. 24 V/m

c. 30 V/m

d. 6 V/m

Answer: (b)

Solution:

$$P_{\text{refracted}} = \frac{96}{100} P_{\text{incident}}$$ $$\implies K_2 A_t^2 = \frac{96}{100} K_1 A_i^2$$ $$\implies \mu_2 A_t^2 = \frac{96}{100} \mu_1 A_i^2$$ $$\implies 1.5 A_t^2 = \frac{96}{100} \times 1 \times (30)^2$$ $$A_t^2 = \frac{64}{100} \times (30)^2 \implies A_t = 24\text{ V/m}$$

68. What is the position and nature of image formed by lens combination shown in figure? ($f_1, f_2$ are focal lengths) [JEE Main 2019]

a. 70 cm from point B at left; virtual

b. 40 cm from point B at right; real

c. $\frac{20}{3}\text{ cm}$ from point B at right , real

d. 70 cm from point B at right, real

Answer: (d)

Solution:

For first lens:

$$\frac{1}{V_1} - \frac{1}{-20} = \frac{1}{5} \implies V_1 = \frac{20}{3}\text{ cm}$$

For second lens:

$$u_2 = \frac{20}{3} - 20 = -\frac{40}{3}\text{ cm}$$ $$\frac{1}{V_2} - \frac{1}{-40/3} = \frac{1}{5} \implies V_2 = 70\text{ cm}$$

69. Apparent depth for point object $x$ in all three cases are $H_1, H_2 \& H_3$ respectively when seen from above given $H = 30\text{ cm}$, $n = 1.5$ & $R = 3\text{ m}$, then [JEE Adv. 2019 P-I]

a. $0.8 < H_2 – H_1 < 0.9$

b. $H_2 > H_1$

c. $H_2 > H_3$

d. $H_3 > H_1$

Answer: (b, c)

Solution:

Case (i): $H = 30\text{ cm}, n = 3/2$

$$H_1 = \frac{H}{n} = 30 \times \frac{2}{3} = 20\text{ cm}$$

Case (ii): $R = 300\text{ cm}$

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$ $$\frac{1}{-H_2} - \frac{3/2}{-30} = \frac{1 - 1.5}{-300}$$ $$H_2 = \frac{600}{29} = 20.684\text{ cm}$$

Case (iii):

$$\frac{1}{-H_3} - \frac{3/2}{-30} = \frac{1 - 1.5}{300}$$ $$H_3 = \frac{600}{31} = 19.354\text{ cm}$$

70. Consider two plane convex lens of same radius of curvature and refractive index $n_1$ and $n_2$ respectively now consider two cases: [JEE Adv. 2019 P-II]

Case (i): When $n_1 = n_2 = n$, then equivalent focal length of lens is $f_0$

Case (ii): When $n_1 = n_2 = n + \Delta n$ then equivalent focal length of lens is $f = f_0 + \Delta f_0$

Then correct options are:

a. If $\Delta n /n > 0$, then $\Delta f_0/ f_0 < 0$

b. $|\Delta f_0/ f_0| < |\Delta n /n |$

c. If $n = 1.5, \Delta n = 10^{-3}$ and $f_0 = 20\text{ cm}$ then $|\Delta f_0| = 0.02\text{ cm}$

d. None of these

Answer: (a, c)

Solution:

$$\frac{1}{f_0} = \frac{2(n-1)}{R} \quad \text{--- (i)}$$ $$\frac{1}{f_0 + \Delta f_0} = \frac{2(n + \Delta n - 1)}{R} \quad \text{--- (ii)}$$

From (i) and (ii), we get:

$$\frac{f_0 + \Delta f_0}{f_0} = \frac{n - 1}{n + \Delta n - 1}$$ $$\frac{\Delta f_0}{f_0} = \frac{-\Delta n}{n + \Delta n - 1}$$

For $n = 1.5, \Delta n = 10^{-3}, f_0 = 20\text{ cm}$:

$$\Delta f_0 = -\frac{20 \times 10^{-3}}{0.5 + 10^{-3}} \approx -0.02\text{ cm} \implies |\Delta f_0| = 0.02\text{ cm}$$

71. In YDSE monochromatic light of wavelength 600 nm incident of slits as shown in figure. [JEE Adv. 2019 P-II] if S1S2 = 3 mm, OP = 11 mm

a. $0.36\frac{\alpha}{\pi} =$ degree then destructive interfaces at point P

b. $0.36\frac{\alpha}{\pi} =$ degree then destructive interfaces at point O

c. If $\alpha = 0$ then constructive interfaces at O

d. Fringe width depends an $\alpha$

Answer: (a, b, c)

Solution:

$$\Delta x = d\sin\alpha + d\sin\theta$$

$\theta, \alpha$: Small angle $\sin\theta \approx \tan\theta \approx \frac{y}{D}$;

$$\Delta x = d\alpha + \frac{dy}{D}$$

(a) $\Delta x = 3\text{ mm} \times \frac{0.36\pi}{180\pi} + \frac{3\text{ mm} \times 11\text{ mm}}{1\text{ m}} = 3900\text{ nm}$

$$600\text{ nm} \times (2n-1) = 3900\text{ nm} \implies 2n-1 = 13 \implies n = 7$$

For $n = 7$, destructive interference happens at point P.

(b) $\Delta x = 3\text{ mm} \times \frac{0.36\pi}{180\pi} + 0 = 600\text{ nm}$

$$600\text{ nm} = n\lambda \implies n = 1\text{ (constructive interference)}$$

(d) Fringe width does not depend on $\alpha$.

Matrix Match/Column Match

72. A simple telescope used to view distant objects has eyepiece and objective lens of focal lengths $f_e$ and $f_0$ respectively. Then [JEE 2006]

Column I

(A) Intensity of light received by lens

(B) Angular magnification

(D) Sharpness of image

Column II

1. Radius of aperture (R)

2. Dispersion of lens

3. focal length

4. spherical aberration

a. A$\rightarrow$1; B$\rightarrow$3; C$\rightarrow$3; D$\rightarrow$1,2,4

b. A$\rightarrow$2; B$\rightarrow$4; C$\rightarrow$1; D$\rightarrow$1,3,4

c. A$\rightarrow$4; B$\rightarrow$3; C$\rightarrow$2; D$\rightarrow$2,3,4

d. A$\rightarrow$3; B$\rightarrow$1; C$\rightarrow$4; D$\rightarrow$1,2,3

Answer: (a)

Solution:

A$\rightarrow$1; B$\rightarrow$3; C$\rightarrow$3; D$\rightarrow$1,2,4

73. An optical component and an object S placed along its optic axis are given in Column I. The distance between the object and the component can be varied. The properties of images are given in Column II. [JEE 2008 P-II]

Column I

(A)

(B)

(C)

(D)

Column II

1. Real image

2. Virtual image

3. Magnified image

4. Image at infinity

a. A $\rightarrow$1,2,3,4; B$\rightarrow$2; C$\rightarrow$ 1,2,3,4; D$\rightarrow$1,2,3,4

b. A $\rightarrow$1,2,3,4; B$\rightarrow$1; C$\rightarrow$ 1,2,3,4; D$\rightarrow$1,2,3,4

c. A $\rightarrow$1,2,3,4; B$\rightarrow$3; C$\rightarrow$ 1,2,3,4; D$\rightarrow$1,2,3,4

d. A $\rightarrow$1,2,3,4; B$\rightarrow$4; C$\rightarrow$ 1,2,3,4; D$\rightarrow$1,2,3,4

Answer: (a)

Solution:

A $\rightarrow$1,2,3,4; B$\rightarrow$2; C$\rightarrow$ 1,2,3,4; D$\rightarrow$1,2,3,4

74. Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away from the slits $S_1$ and $S_2$. In each of these cases $S_1P_0 = S_2P_0$, $S_1P_1 - S_2P_1 = \lambda/4$ and $S_1P_2 - S_2P_2 = \lambda/3$, where $\lambda$ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index $\mu$ and thickness $t$ is pasted on slit $S_2$. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by $\delta(P)$ and the intensity by $I(P)$. Match each situation given in Column I with the statement(s) in Column II valid for that situation. [JEE 2009 P-II]

Column I

(A)

(B) $(\mu-1)t = \lambda/4$

(D) $(\mu-1)t = 3\lambda/4$

Column II

1. $\delta(P_0) = 0$

2. $\delta(P_1) = 0$

3. $I(P_1) = 0$

4. $I(P_0) > I(P_1)$

5. $I(P_2) > I(P_1)$

a. A$\rightarrow$1,4; B$\rightarrow$2; C$\rightarrow$5; D$\rightarrow$3,4,5

b. A$\rightarrow$3,4; B$\rightarrow$4; C$\rightarrow$5; D$\rightarrow$1,2,3

c. A$\rightarrow$2,5; B$\rightarrow$1; C$\rightarrow$4; D$\rightarrow$2,3,4

d. A$\rightarrow$4,5; B$\rightarrow$3; C$\rightarrow$2; D$\rightarrow$1,3,4

Answer: (a)

Solution:

A$\rightarrow$1,4; B$\rightarrow$2; C$\rightarrow$5; D$\rightarrow$3,4,5

75. Two transparent media of refractive indices $\mu_1$ and $\mu_3$ have a solid lens shaped transparent material of refractive index $\mu_2$ between them as shown in figures in Column II. A ray traversing these media is also shown in the figures. In Column I different relationships between $\mu_1, \mu_2$ and $\mu_3$ are given. Match them to the ray diagram shown in Column II. [JEE 2010 P-II]

Column I

(A) $\mu_1 < \mu_2$

(B) $\mu_1 > \mu_2$

(D) $\mu_2 > \mu_3$

Column II

a. A$\rightarrow$1,3; B$\rightarrow$2,4,1; C$\rightarrow$1,3,5; D$\rightarrow$2,4

b. A$\rightarrow$2,4; B$\rightarrow$1,2,3; C$\rightarrow$2,3,4; D$\rightarrow$4,5

c. A$\rightarrow$3,4; B$\rightarrow$2,4,5; C$\rightarrow$1,2,3; D$\rightarrow$1,3

d. A$\rightarrow$4,5; B$\rightarrow$3,4,5; C$\rightarrow$2,4,5; D$\rightarrow$3,4

Answer: (a)

Solution:

A$\rightarrow$1,3; B$\rightarrow$2,4,1; C$\rightarrow$1,3,5; D$\rightarrow$2,4

76. A right angled prism of refractive index $\mu_1$ is placed in a rectangular block of refractive index $\mu_2$, which is surrounded by a medium of refractive index $\mu_3$, as shown in the figure. A ray of light ‘e’ enters the rectangular block at normal incidence. Depending upon the relationships between $\mu_1, \mu_2$ and $\mu_3$, it takes one of the four possible paths ‘ef’, ‘eg’, ‘eh’, or ‘ei’.

Match the paths in Column I with conditions of refractive indices in Column II: [JEE Adv. 2013 P-II]

Column I

(A) ef $\rightarrow$

(B) eg $\rightarrow$

(D) ei $\rightarrow$

Column II

1. $\mu_1 > \mu_2$

2. $\mu_2 > \mu_1$ and $\mu_2 > \mu_3$

3. $\mu_1 = \mu_2$

4. $\mu_1 < \mu_2 < 2\mu_1$ and $\mu_2 > \mu_3$

a. A$\rightarrow$ 2; B$\rightarrow$ 3; C$\rightarrow$ 4; D$\rightarrow$ 1

b. A$\rightarrow$ 1; B$\rightarrow$ 2; C$\rightarrow$ 3; D$\rightarrow$ 4

c. A$\rightarrow$ 3; B$\rightarrow$ 1; C$\rightarrow$ 2; D$\rightarrow$ 4

d. A$\rightarrow$ 3; B$\rightarrow$ 4; C$\rightarrow$ 2; D$\rightarrow$ 1

Answer: (a)

Solution:

(A) $\mu_2 > \mu_1$ ... (Towards normal); $\mu_2 > \mu_3$ ... (Away from normal)

(B) $\mu_1 = \mu_2$ ... (No change in path)

$$\mu_1 \sin r = \mu_2 \times \frac{1}{\sqrt{2}} \implies \sin r = \frac{\mu_2}{\sqrt{2}\mu_1}$$

Since $\sin r < 1 \implies \mu_2 < \sqrt{2}\mu_1$

(D) For TIR:

$$45^\circ > C \implies \sin 45^\circ > \sin C \implies \frac{1}{\sqrt{2}} > \frac{\mu_2}{\mu_1} \implies \mu_1 > \sqrt{2}\mu_2$$

Statement or Assertion and Reason

77. Statement-I: The formula connecting u, v and f for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature. And Statement-II: Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces. [JEE 2007 P-I]

a. Statement-I is True, Statement-II is True; Statement -II is a correct explanation for Statement-I.

b. Statement-I is True, Statement-II is True; Statement -II is not a correct explanation for Statement-I.

c. Statement -I is True, Statement-II is False.

d. Statement -I is False, Statement-II is True.

Answer: (c)

Paragraphs

Paragraph for Question Nos. 78 to 80

The figure shows a surface XY separating two transparent media, medium-1 and medium -2. The lines ab and cd represent wave fronts of a light wave travelling in medium-1 and incident on XY. The lines ef and gh represent wave fronts of the light wave in medium-2 after refraction. [JEE 2007 P-II]

78. Light travels as a

a. parallel beam in each medium

b. convergent beam in each medium

c. divergent beam in each medium

d. divergent beam in one medium and convergent beam in the other medium.

Answer: (a)

79. The phases of the light wave at c, d, e and f are $\phi_c, \phi_d, \phi_e$ and $\phi_f$ respectively. It is given that $\phi_c \neq \phi_f$

a. $\phi_c$ cannot be equal to $\phi_d$

b. $\phi_d$ can be equal to $\phi_e$

c. $(\phi_d - \phi_f)$ is equal to $(\phi_c - \phi_e)$

d. $(\phi_d - \phi_c)$ is not equal to $(\phi_f - \phi_e)$

Answer: (c)

80. Speed of the light is

a. the same in medium-1 and medium-2

b. larger in medium-1 than in medium-2

c. larger in medium-2 than in medium-1

d. different at b and d

Answer: (b)

Paragraph for Question Nos. 81 to 82

Most materials have the refractive index, $n > 1$. So, when a light ray from air enters a naturally occurring material, then by Snell’s law, $\frac{\sin\theta_1}{\sin\theta_2} = \frac{n_2}{n_1}$ it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, $n = \frac{c}{v} = \pm \sqrt{\varepsilon_r \mu_r}$ where c is the speed of electromagnetic waves in vacuum, v its speeds in the medium, $\varepsilon_r$ and $\mu_r$ are the relative permittivity and permeability of the medium respectively. In normal materials, both $\varepsilon_r$ and $\mu_r$ are positive, implying positive n for the medium. When both $\varepsilon_r$ and $\mu_r$ are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called metamaterials. They exhibit significantly different optical behavior, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials. [JEE 2012 P-II]

81. Choose the correct statement.

a. The speed of light in the meta-material is $v = c|n|$

b. The speed of light in the meta-material is $v = \frac{c}{|n|}$

c. The speed of light in the meta-material is $v = c$.

d. The wavelength of the light in the meta–material ($\lambda_m$) is given by $\lambda_m = \lambda_{\text{air}}|n|$, where $\lambda_{\text{air}}$ is the wavelength of the light in air

Answer: (b)

Solution:

$\eta = \frac{c}{v} \implies v = \frac{c}{\eta}$

$\eta$ will be $(-)\text{ve}$ for metamaterials but $|\eta|$ is $(+)\text{ve}$, so $v = \frac{c}{|\eta|}$.

82. For light incident from air on a meta-material, the appropriate ray diagram is. [JEE 2012 P-II]

Answer: (a)

Solution:

For minimum deviation $\angle r_1 = \angle r_2 = \frac{B}{2} \implies \frac{\sin i}{\sin 30^\circ} = \sqrt{3} \implies i = 60^\circ$

Prism DCE should be rotated about C in anticlockwise direction through $60^\circ$ so that the final emergent ray is parallel to the incident ray and angle of deviation is zero (minimum).

Paragraph for Question Nos. 83 to 84

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index $n_1$ surrounded by a medium of lower refractive index $n_2$. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media $n_1$ and $n_2$ as shown in the figure. All rays with the angle of incidence i less than a particular value $i_m$ are confined in the medium of refractive index $n_1$. The numerical aperture (NA) of the structure is defined as $\sin i_m$. [JEE Adv. 2015 P-II]

83. For two structures namely $S_1$ with $n_1 = 45/4$ and $n_2 = 3/2$, and $S_2$ with $n_1 = 8/5$ and $n_2 = 7/5$ and taking the refractive index of water to be $4/3$ and that of air to be 1, the correct option(s) is(are)

a. NA of $S_1$ immersed in water is the same as that of $S_2$ immersed in a liquid of refractive index $\frac{16}{3\sqrt{15}}$

b. NA of $S_1$ immersed in liquid of refractive index $\frac{6}{\sqrt{15}}$ is the same as that of $S_2$ immersed in water

c. NA of $S_1$ placed in air is the same as that of $S_2$ immersed in liquid of refractive index $\frac{4}{\sqrt{15}}$

d. NA of $S_1$ placed in air is the same as that of $S_2$ placed in water

Answer: (c)

Solution:

$$\frac{\sin\theta_1}{\sin\theta_2} = \frac{\eta_2}{\eta_1}$$ $$\therefore \sin\theta_1 = \sqrt{1 - \frac{\eta_2^2}{\eta_1^2}}$$ $$\Rightarrow \eta \sin\theta_1 = \sqrt{\eta_1^2 - \eta_2^2}$$

$\theta_2$ will be $(-)\text{ve}$ w.r.t. to normal.

$\sin i_m = \frac{\sqrt{n_1^2 - n_2^2}}{n_m}$

Putting values, we get, correct option as C.

84. If two structures of same cross-sectional area, but different numerical apertures $\text{NA}_1$ and $\text{NA}_2$ ($\text{NA}_2 < \text{NA}_1$) are joined longitudinally, the numerical aperture of the combined structure is:

a. $\frac{\text{NA}_1 \text{NA}_2}{\text{NA}_1 + \text{NA}_2}$

b. $\text{NA}_1 + \text{NA}_2$

c. $\text{NA}_1$

d. $\text{NA}_2$

Answer: (d)

Solution:

For total internal reflection to take place in both structures, the numerical aperture should be the least one for the combined structure and hence, correct option is (d).

Integer and Subjective

85. Two identical prisms of refractive index $\sqrt{3}$ are kept as shown in the figure. A light ray strikes the first prism at face AB. Find,

(a) The angle of incidence, so that the emergent ray from the first prism has minimum deviation.

(b) Through what angle the prism DCE should be rotated about C so that the final emergent ray also has minimum deviation. [JEE 2005]

Solution:

(a) For minimum deviation $\angle r_1 = \angle r_2 = \frac{B}{2} \implies \frac{\sin i}{\sin 30^\circ} = \sqrt{3} \implies i = 60^\circ$

(b) Prism DCE should be rotated about C in anticlockwise direction through $60^\circ$ so that the final emergent ray is parallel to the incident ray and angle of deviation is zero (minimum).

86. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\frac{m_{25}}{m_{50}}$ is [JEE 2010 P-I]

Answer: 6

Solution:

$$m = \frac{f}{f + u}$$ $$\frac{m_{25}}{m_{50}} = \frac{f / (f + u_1)}{f / (f + u_2)} = \frac{f + u_2}{f + u_1} = \frac{20 + (-50)}{20 + (-25)} = \frac{-30}{-5} = 6$$

87. A large glass slab ($\mu = 5/3$) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R? [JEE 2010 P-II]

Answer: 6

Solution:

$$\sin\theta_c = \frac{1}{\mu} = \frac{3}{5}$$ $$\tan\theta_c = \frac{R}{h} \implies \frac{3}{4} = \frac{R}{8} \implies R = 6\text{ cm}$$

88. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from $\frac{25}{3}\text{ m}$ to $\frac{50}{7}\text{ m}$ in 30 seconds. What is the speed of the object in km per hour? [JEE 2010 P-II]

Answer: 3

Solution:

For $v_1 = \frac{50}{7}\text{ m}$, $u_1 = -25\text{ m}$

For $v_2 = \frac{25}{3}\text{ m}$, $u_2 = -50\text{ m}$

$$\text{Speed of object} = \frac{50 - 25}{30} = \frac{25}{30}\text{ m/s} = \frac{25}{30} \times \frac{18}{5} = 3\text{ kmph}$$

89. Water (with refractive index $\frac{4}{3}$) in a tank is 18 cm deep. Oil of refractive index $\frac{7}{4}$ lies on water making a convex surface of radius of curvature $R = 6\text{ cm}$ as shown. Consider oil to act as a thin lens. An object ‘S’ is placed 24 cm above water surface. The location of its image is at ‘x’ cm above the bottom of the tank. Then ‘x’ is [JEE 2011 P-II]

Answer: 2

Solution:

$$\frac{7/4}{V_1} - \frac{1}{-24} = \frac{7/4 - 1}{6} \implies V_1 = 21\text{ cm}$$ $$\frac{4/3}{V_2} - \frac{7/4}{21} = 0 \implies V_2 = 16\text{ cm}$$ $$x = 18 - 16 = 2\text{ cm}$$

90. Consider a concave mirror and a convex lens (refractive index $= 1.5$) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index $= 1$) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification $M_1$. When the set-up is kept in a medium of refractive index 7/6, the magnification becomes $M_2$. The magnitude $\left|\frac{M_1}{M_2}\right|$ is: [JEE Adv. 2015 P-I]

Answer: 7

Solution:

Image by mirror is formed at 30 cm from mirror at its right and finally by the combination it is formed at 20 cm on right of the lens. So in air medium, magnification by lens is unity. In second medium, focal length of the lens is given by,

$$\mu = \frac{7}{6}$$ $$\frac{1}{f'} = \left(\frac{1.5}{7/6} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \times \frac{2}{7}$$ $$\implies f' = \frac{35}{2}\text{ cm}$$

So in second medium, final image is formed at 140 cm to the right of the lens. Second medium does not change the magnification by mirror. So

$$\left|\frac{M_1}{M_2}\right| = 7$$

91. A monochromatic beam of light is incident at $60^\circ$ on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle $\theta(n)$ with the normal (see the figure). For $n = \sqrt{3}$ the value of $\theta$ is $60^\circ$ and $\frac{d\theta}{dn} = m$. The value of m is: [JEE Adv. 2015 P-II]

Answer: 2

Solution:

Snell’s Law on 1st surface:

$$1 \cdot \sin 60^\circ = n \sin r_1 \implies \frac{\sqrt{3}}{2} = n \sin r_1 \quad \text{...(i)}$$ $$\implies \cos r_1 = \sqrt{1 - \frac{3}{4n^2}} = \frac{\sqrt{4n^2-3}}{2n}$$ $$r_1 + r_2 = 60^\circ \quad \text{...(ii)}$$

Snell’s Law on 2nd surface: $n \sin r_2 = \sin\theta$

Using equation (i) and (ii): $n \sin(60^\circ - r_1) = \sin\theta$

$$n\left[\frac{\sqrt{3}}{2}\cos r_1 - \frac{1}{2}\sin r_1\right] = \sin\theta$$ $$\frac{\sqrt{3}}{4}\sqrt{4n^2-3} - \frac{\sqrt{3}}{4} = \sin\theta$$

Differentiating w.r.t $n$:

$$\cos\theta \frac{d\theta}{dn} = \frac{\sqrt{3}}{4} \cdot \frac{8n}{2\sqrt{4n^2-3}}$$

For $\theta = 60^\circ$ and $n = \sqrt{3} \implies \frac{d\theta}{dn} = 2$, so $m = 2$.

92. A monochromatic light is travelling in a medium of refractive index $n = 1.6$. It enters a stack of glass layers from the bottom side at an angle $\theta = 30^\circ$. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as $n_m = n - m\Delta n$, where $n_m$ is the refractive index of the $m$-th slab and $\Delta n = 0.1$ (see the figure). The ray is refracted out parallel to the interface between the $(m - 1)$-th and $m$-th slabs from the right side of the stack. What is the value of m? [JEE Adv. 2017 P-I]

Answer: 8

Solution:

For refracted ray, to emerge parallel to the interface:

$$n\sin\theta = (n - m\Delta n)\sin 90^\circ$$ $$1.6 \sin 30^\circ = (1.6 - m \times 0.1)$$ $$0.8 = 1.6 - 0.1m \implies 0.1m = 0.8 \implies m = 8$$

93. Sunlight of intensity $1.3\text{ kW m}^{-2}$ is incident normally on a thin convex lens of focal length 20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, $\text{kW m}^{-2}$ at a distance 22 cm from the lens on the other side is ______. [JEE Adv. 2018 P-I]

Answer: 130

Solution:

Power incident, $P = I_1 A_1$

Intensity on screen:

$$I_2 = \frac{P}{A_2} = I_1 \left(\frac{A_1}{A_2}\right) = I_1 \left(\frac{20}{2}\right)^2 = 100 I_1$$ $$I_2 = 100 \times 1.3 = 130\text{ kW m}^{-2}$$

94. An optical bench, to measure the focal length of lens, is 1.5 m long and on the bench marks are with spacing $1.4\text{ cm}$. Now a lens is placed at 75 cm and pin type object is placed at 45 cm marks on the bench. If its image is formed at 135 cm find maximum possible error in calculation of focal length. [JEE Adv. 2019 P-I]

Answer: 1.38 (or 1.39)

Solution:

$$u = -(x_2 - x_1) = -(75 - 45) = -30\text{ cm}$$ $$\Delta u = \Delta x_2 + \Delta x_1 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\text{ cm}$$ $$v = x_3 - x_2 = 135 - 75 = 60\text{ cm}$$ $$\Delta v = \Delta x_3 + \Delta x_2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\text{ cm}$$ $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{60} + \frac{1}{30} = \frac{1}{f} \implies f = 20\text{ cm}$$ $$\frac{df}{f^2} = \frac{dv}{v^2} + \frac{du}{u^2}$$ $$df = f^2 \left[\frac{dv}{v^2} + \frac{du}{u^2}\right] = (20)^2 \left[\frac{1/2}{60^2} + \frac{1/2}{30^2}\right] = 400 \left[\frac{1}{7200} + \frac{1}{1800}\right] = \frac{50}{36} \approx 1.38\text{ or }1.39\text{ cm}$$

95. A prism is shown in the figure with angle 75º and refractive index $\sqrt{3}$. A light ray incidents on a surface at incident angle $\theta$. Other face is coated with a medium of refractive index n. For $\theta \le 60^\circ$ ray suffers total internal reflection find value of $n^2$. [JEE Adv. 2019 P-II]

Answer: 1.50

Solution:

For T.I.R at coating:

$$\sin c = \frac{n}{\sqrt{3}}$$

Applying Snell’s law at first surface:

$$\sin\theta = \sqrt{3} \sin(75^\circ - c)$$

For limiting condition, at $\theta = 60^\circ$:

$$\sin 60^\circ = \sqrt{3} \sin(75^\circ - c)$$ $$\frac{\sqrt{3}}{2} = \sqrt{3} \sin(75^\circ - c)$$ $$\frac{1}{2} = \sin(75^\circ - c) \implies \sin 30^\circ = \sin(75^\circ - c)$$ $$30^\circ = 75^\circ - c \implies c = 45^\circ$$ $$\sin 45^\circ = \frac{n}{\sqrt{3}} \implies \frac{1}{\sqrt{2}} = \frac{n}{\sqrt{3}} \implies n = \sqrt{\frac{3}{2}}$$ $$n^2 = \frac{3}{2} = 1.50$$

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