JEE Physics Ray Optics and Optical Instruments MCQs Set 15

Answer: (a) 120 cm

Explanation:
1. Focal length of each lens: \( |f|=\frac{R}{\mu-1}=\frac{30}{0.75}=40\ \text{cm} \), so \( f_1=-40\ \text{cm} \) and \( f_2=+40\ \text{cm} \).
2. The concave lens forms a virtual image at 40 cm to its left (its focal point), which is 80 cm from the second lens: \( u_2=-80\ \text{cm} \).
3. \( \frac1{v_2}=\frac1{40}-\frac1{80}=\frac1{80}\Rightarrow v_2=80\ \text{cm} \) from the convex lens.
4. Distance from the concave lens \( =40+80=120\ \text{cm} \).

Teacher's Note:
1. Treat the image formed by the first lens as the object for the second lens.
2. Parallel light after a diverging lens appears to come from its focal point.

Answer: (b) Chromatic aberration

Explanation:
1. The refractive index of the lens material is different for different colours, so each colour has its own focal length.
2. This colour-dependent focusing is chromatic aberration.

Teacher's Note:
1. Violet light is refracted most and focuses closest to the lens.
2. Spherical aberration, in contrast, occurs for a single colour with rays at different heights from the axis.

Answer: (a) 30 cm

Explanation:
1. A plane mirror forms an image as far behind it as the object is in front, so the object for the mirror (the image from the lens) is 5 cm in front of the mirror, i.e. \( 20-5=15\ \text{cm} \) from the lens.
2. Lens formula with \( v=+15\ \text{cm},\ f=+10\ \text{cm} \): \( \frac1u=\frac1v-\frac1f=\frac1{15}-\frac1{10}=-\frac1{30} \).
3. \( u=-30\ \text{cm} \), so the object is 30 cm from the lens.

Teacher's Note:
1. Find the position of the intermediate image formed by the lens using the mirror property.
2. Use the sign convention 1/v - 1/u = 1/f.

Answer: (a) B and D only

Explanation:
1. A plane mirror forms a virtual image, erect, of the same size as the object, and laterally inverted.
2. So B and D are correct; A (real) and C (smaller) are incorrect.

Teacher's Note:
1. The image seems to be behind the mirror, so it cannot be real.
2. Left and right are interchanged in a plane mirror image.

Answer: (a) 54 cm

Explanation:
1. \( \frac1{f}=(\mu-1)\left(\frac1{R_1}-\frac1{R_2}\right) \): in air \( \frac1{18}=0.5\,K \), so \( K=\frac19 \).
2. In water \( \frac1{f_w}=\left(\frac{1.5}{4/3}-1\right)K=0.125\times\frac19\Rightarrow f_w=72\ \text{cm} \).
3. Change in focal length \( =72-18=54\ \text{cm} \).

Teacher's Note:
1. The effective refractive index is the ratio of the lens material to the surrounding medium.
2. A lens loses much of its power when immersed in a liquid of similar index.

Answer: (a) 0.52 cm

Explanation:
1. Critical angle: \( \sin i_c=\frac1{\sqrt2}\Rightarrow i=45^\circ \).
2. Snell's law: \( \sin r=\frac{\sin45^\circ}{\sqrt2}=\frac12\Rightarrow r=30^\circ \).
3. Lateral displacement \( d=\frac{t\sin(i-r)}{\cos r}=\frac{\sqrt3\times0.26}{\sqrt3/2}=0.52\ \text{cm} \).

Teacher's Note:
1. The lateral shift depends on the thickness, the angle of incidence and the angle of refraction.
2. i - r = 15° here, which is why sin 15° is given.

Answer: (d) C and D only

Explanation:
1. A is false: the speed of light in vacuum is the same in all directions.
2. B is false: in a medium (dispersion) the speed depends on wavelength.
3. C is true: the speed of light does not depend on the motion of the source.
4. D is true: the speed in a medium does not depend on intensity (for ordinary intensities).

Teacher's Note:
1. The constancy of the speed of light in vacuum is a postulate of special relativity.
2. Dispersion arises because the refractive index depends on wavelength.

Answer: (b) Increase the refractive index of the medium between the object and the objective lens

Explanation:
1. Resolving power of a microscope \( =\frac{2\mu\sin\theta}{1.22\lambda} \).
2. It increases when the refractive index \( \mu \) of the medium between the object and the objective increases (oil immersion), when the wavelength decreases, and when the aperture of the objective increases.

Teacher's Note:
1. Oil-immersion objectives use a liquid of high refractive index for this reason.
2. The eyepiece only magnifies; it does not improve resolution.

Answer: (a) 41

Explanation:
1. One main scale division \( =\frac{1}{20}\ \text{cm}=0.5\ \text{mm} \), so the least count \( =\frac{0.5}{50}=0.01\ \text{mm} \).
2. Refractive index \( \mu=\frac{5.25}{5.00}=1.05 \).
3. \( \Delta\mu=\mu\left(\frac{0.01}{5.25}+\frac{0.01}{5.00}\right)=1.05\times3.905\times10^{-3}=4.1\times10^{-3} \), so \( \frac{x}{10}=4.1 \) and x = 41.

Teacher's Note:
1. The least count of a vernier is one main scale division divided by the number of vernier divisions.
2. The fractional errors of the two measurements add.

Answer: (a) 32

Explanation:
1. Magnitudes: \( \frac1f=\frac1{40}+\frac1{120}\Rightarrow f=30\ \text{cm} \).
2. Least count \( =\frac1{20}=0.05\ \text{cm} \).
3. \( \frac{\Delta f}{f^2}=\frac{\Delta u}{u^2}+\frac{\Delta v}{v^2}\Rightarrow\Delta f=900\times0.05\left(\frac1{1600}+\frac1{14400}\right)=\frac{1}{32}\ \text{cm} \), so K = 32.

Teacher's Note:
1. Differentiate 1/f = 1/u + 1/v and add the absolute errors.
2. The error in each distance is taken equal to the least count.

Answer: (d) 50 cm

Explanation:
1. The reading glass has focal length \( f=\frac{100}{2}=50\ \text{cm} \) and is used to read a book held at the normal distance of 25 cm.
2. Lens formula with \( u=-25\ \text{cm} \): \( \frac1v=\frac1f+\frac1u=\frac1{50}-\frac1{25}=-\frac1{50} \), so \( v=-50\ \text{cm} \).
3. The image forms at 50 cm, which is the person's least distance of distinct vision.

Teacher's Note:
1. The reading glass places the book's image at the person's near point.
2. The distance glasses (-1 D) only correct the far point.

Answer: (d) 4.5°

Explanation:
1. No (mean) deviation: \( (\mu_1-1)A_1=(\mu_2-1)A_2 \).
2. \( A_2=\frac{0.54\times6^\circ}{0.72}=4.5^\circ \).

Teacher's Note:
1. The two prisms are placed with their bases opposite, so their deviations cancel.
2. δ = (μ - 1)A for a thin prism.

Answer: (a) 5

Explanation:
1. Refractive index \( n=\frac{c}{v}=\frac1{0.2}=5 \).
2. Relative permittivity \( \varepsilon_r=\frac{n^2}{\mu_r}=25 \).
3. Ratio \( \frac{\varepsilon_r}{n}=\frac{25}{5}=5 \), so x = 5.

Teacher's Note:
1. Maxwell's relation: n = √(μr εr).
2. With μr = 1, εr = n².

Answer: (a) 75 cm

Explanation:
1. Apparent depth \( =\frac{h}{\mu} \), so the apparent rise of the object is \( h-\frac h\mu=h\left(1-\frac35\right)=0.4h \).
2. \( 0.4h=30\Rightarrow h=75\ \text{cm} \).

Teacher's Note:
1. The microscope must move by the same amount as the apparent shift of the object.
2. Real depth minus apparent depth is the shift.

Answer: (a) 32

Explanation:
1. The ends of the rod are at 35 cm and 45 cm from the pole.
2. Mirror formula (magnitudes, real image): \( v=\frac{uf}{u-f} \).
3. For u = 35: \( v_1=\frac{35\times20}{15}=\frac{140}{3}\ \text{cm} \). For u = 45: \( v_2=\frac{45\times20}{25}=36\ \text{cm} \).
4. Image length \( =\frac{140}3-36=\frac{32}3\ \text{cm} \), so x = 32.

Teacher's Note:
1. Find the images of the two end points separately.
2. The rod lies along the axis, so the image length is the difference of the image distances.

Answer: (c) 160 cm

Explanation:
1. Focal length \( f=-20\ \text{cm} \).
2. Object A: \( \frac1v=\frac1f-\frac1u=-\frac1{20}+\frac1{15}=\frac1{60}\Rightarrow v=+60\ \text{cm} \) (virtual image behind the mirror).
3. Object B: \( \frac1v=-\frac1{20}+\frac1{25}=-\frac1{100}\Rightarrow v=-100\ \text{cm} \) (real image in front).
4. Distance between the images \( =60+100=160\ \text{cm} \).

Teacher's Note:
1. The object at 15 cm lies inside the focal length, so its image is virtual and behind the mirror.
2. Add the distances because the images are on opposite sides of the mirror.