JEE Physics Ray Optics and Optical Instruments MCQs Set 28

Answer: (a) 24 V/m

Explanation:
1. 96% of the intensity is transmitted. Intensity \( \propto n E^2 \), so \( 1.5E_t^2=0.96\times30^2 \).
2. \( E_t^2=576\Rightarrow E_t=24 \) V/m.

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (b) 3.1 cm

Explanation:
1. \( \dfrac{1.34}{v}-\dfrac1\infty=\dfrac{1.34-1}{7.8} \).
2. \( v=\dfrac{1.34\times7.8}{0.34}=30.7 \) mm \( \approx3.1 \) cm.

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (c) \( \dfrac3{\sqrt5} \)

Explanation:
1. The angle of refraction r in the liquid satisfies \( \sin r\le\dfrac1\mu \) (i up to \( 90^\circ \)). Brewster's angle at the liquid-glass interface: \( \tan\theta_B=\dfrac{1.5}\mu \).
2. Never polarised means \( \tan\theta_B\gt \tan r_{max}=\dfrac1{\sqrt{\mu^2-1}} \): \( 2.25(\mu^2-1)\gt \mu^2\Rightarrow\mu^2\gt \dfrac95 \), so \( \mu_{min}=\dfrac3{\sqrt5} \).

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (d) \( 30\ \mu\text{m} \)

Explanation:
1. \( \Delta\theta=\dfrac{1.22\lambda}D=\dfrac{1.22\times5\times10^{-7}}{5\times10^{-3}}=1.22\times10^{-4} \) rad.
2. Separation \( =25\ \text{cm}\times1.22\times10^{-4}\approx30\ \mu m \).

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (a) bends upwards

Explanation:
1. The wavefront moves faster where the index is lower, i.e. near the ground.
2. The lower part of the wavefront runs ahead, tilting the wavefront and bending the beam upwards (as in a mirage).

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (b) \( \dfrac13 \)

Explanation:
1. After rotation through \( 30^\circ \): beam A gives \( I_A\cos^230^\circ \) and beam B gives \( I_B\sin^230^\circ \).
2. Equal: \( \dfrac34I_A=\dfrac14I_B\Rightarrow\dfrac{I_A}{I_B}=\dfrac13 \).

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (c) 30 cm

Explanation:
1. \( \mu=\dfrac c v=1.5 \). The sagitta is 3 mm with half-chord 30 mm, so \( R=\dfrac{30^2+3^2}{2\times3}\approx151.5 \) mm \( \approx15 \) cm.
2. \( f=\dfrac R{\mu-1}=\dfrac{15}{0.5}=30 \) cm.

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (d) real and at a distance of 16 cm from the mirror

Explanation:
1. First pass: \( u=-30 \), \( v=30 \) cm, i.e. 20 cm behind the mirror (virtual object for the mirror). The mirror makes a real image 20 cm in front of it, i.e. 10 cm beyond the lens.
2. For the lens on the return pass this is a virtual object, \( u=+10 \): \( \dfrac1v=\dfrac1{15}+\dfrac1{10}=\dfrac16\Rightarrow v=6 \) cm from the lens, i.e. \( 10+6=16 \) cm from the mirror: a real image in front of the mirror.

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (a) \( \sin^{-1}\left(\dfrac18\right) \)

Explanation:
1. Snell's law across parallel layers: \( n_0\sin\theta=n_{IV}\sin\theta_{IV} \).
2. Just missing Region IV means \( \theta_{IV}=90^\circ \): \( n_0\sin\theta=\dfrac{n_0}8\Rightarrow\sin\theta=\dfrac18 \).

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (b) 20 cm

Explanation:
1. The apparent depth of the object is \( \dfrac{33.25}{1.33}=25 \) cm, so it appears 40 cm from the mirror.
2. Image at the same place means the object is at the centre of curvature: \( R=40 \) cm and \( f=20 \) cm.

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (c) a convex lens of focal length less than 0.25 m

Explanation:
1. Only a convex lens (or concave mirror) forms real images on a screen. For object-screen distance \( D=u+v \) and \( f=\dfrac{uv}{u+v} \), with \( uv\le\dfrac{D^2}4 \).
2. So \( f\le\dfrac D4=0.25 \) m; for a diminished image \( f\lt 0.25 \) m.

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (d) 130

Explanation:
1. The beam converges to the focus at 20 cm; at 22 cm it has diverged again to a spot whose linear size is \( \dfrac{2}{20}=\dfrac1{10} \) of the aperture.
2. Area ratio \( =\dfrac1{100} \), so the intensity is \( 1.3\times100=130 \) kW m\(^{-2}\).

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (a) 70 cm

Explanation:
1. Rod S1: light from P (in glass) goes into air: \( \dfrac1v-\dfrac{1.5}{-50}=\dfrac{1-1.5}{-10}\Rightarrow\dfrac1v=0.05-0.03=0.02 \), \( v=50 \) cm in air beyond the S1 face.
2. For S2 the rays inside are parallel: \( \dfrac{1.5}\infty-\dfrac1u=\dfrac{0.5}{10}\Rightarrow u=-20 \) cm, so the point is 20 cm before S2. Hence \( d=50+20=70 \) cm.

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.

Answer: (b) 10 cm

Explanation:
1. The lens alone forms the image at \( v=60 \) cm, which is 20 cm in front of the mirror. The image stays there only if the converging rays strike the mirror normally, i.e. 20 cm is the radius of curvature.
2. So \( f=10 \) cm, and a concave mirror gives a virtual image only for objects inside the focal length, at most 10 cm.

Teacher's Note:
1. Draw a rough ray diagram, write the lens or mirror formula for each stage with signs, and use the image of one stage as the object of the next.
2. Check limiting cases (object at focus or at centre of curvature) to confirm the result.