JEE Physics Ray Optics and Optical Instruments MCQs Set 32

Answer: (c) 3°

Explanation:
1. Zero net deviation needs (μ₁ – 1)A₁ = (μ₂ – 1)A₂ (the prisms are placed in opposite orientation).
2. \( 0.54 \times 4 = 0.72 \) × A₂, so A₂ = 3°.

Teacher's Note:
1. Deviation of a thin prism (= μ – 1)A.

Answer: (c) 2

Explanation:
1. The source is at a distance \( 2R \) from the exit surface. Refraction from glass to air at a convex surface with the image at \( \infty \): \( \frac{1}{\infty} - \frac{\mu}{-2R} = \frac{1-\mu}{-R} \).
2. So μ/2R = (μ – 1)/R, giving \( \mu = 2 \).

Teacher's Note:
1. Parallel emergent beam means the image is at \( \infty \).

Answer: (c) 60°

Explanation:
1. Inside, the ray meets the second face at 60°. The critical angle is \( \sin^{-1}(1/1.5) \approx 41.8^{\circ} \), so there is total internal reflection.
2. After reflection, the ray travels perpendicular to the base and leaves it undeviated.
3. Net turning of the ray, from its incident direction to the final direction, is 60°.

Teacher's Note:
1. Check for TIR before applying the prism formula.

Answer: (d) 30 cm and 6 cm respectively

Explanation:
1. For the final image at \( \infty \): f_o + f_e = 36 cm and f_o/f_e = 5.
2. So f_e = 6 cm and f_o = 30 cm.

Teacher's Note:
1. Normal adjustment: tube length = f_o + f_e.

Answer: (a) if the rays incident on the mirror are converging

Explanation:
1. A plane mirror forms a virtual image of a real object (diverging rays).
2. If the incident rays are converging, they meet after reflection in front of the mirror, so a real image is formed (virtual object).

Teacher's Note:
1. A real image needs the reflected rays to converge.

Answer: (c) 60 cm

Explanation:
1. Lens maker's formula: \( 1/f \propto (\mu_l/\mu_m - 1) \).
2. In air: 1.5 – \( 1 = 0.5 \). In the medium: 1.5/(4/3) – \( 1 = 0.125 \).
3. The ratio is 4, so \( f = 4 \times 15 = 60 \) cm.

Teacher's Note:
1. Power of a lens depends on the surrounding medium.

Answer: (b) is formed at the least distance of distinct vision

Explanation:
1. The eyepiece magnification is greatest, 1 + D/f_e, when the final image is at the near point \( (D) \).

Teacher's Note:
1. At \( \infty \): D/f_e. At \( D \): 1 + D/f_e.

Answer: (c) 0.5 m

Explanation:
1. \( u \) = –f. \( 1/v = 1/f \) – \( 1/u = 1/f + 1/f = 2/f \), so \( v = f/2 \).
2. Magnification \( m \) = –v/u = 1/2, so the image height is \( 0.5 m \).

Teacher's Note:
1. A convex mirror always gives a smaller, erect image.

Answer: (a) (i) and (iii) only

Explanation:
1. For an observer in the denser medium (water), the apparent height of an object in air is μ times its real height, so the bird appears farther away.
2. The apparent speed is also μ times the real speed, so it appears faster.

Teacher's Note:
1. Rarer to denser: objects appear farther.
2. Apparent height = μ × real height.

Answer: (b) 2h/μ

Explanation:
1. The image in the plane mirror is \( h \) below the mirror, so the real distance between \( P \) and its image is \( 2h \).
2. Seen from above through the liquid, all depths along the vertical appear reduced by the factor μ.
3. Apparent distance = \( 2h/\mu \).

Teacher's Note:
1. Apparent depth = real depth/μ for near-normal viewing.

Answer: (a) a virtual image of S will be formed at a finite distance

Explanation:
1. In normal adjustment, the image from the objective lies at the focus of the eyepiece.
2. If the tube is shortened, this image lies inside the focal length of the eyepiece, which then acts as a magnifier forming a virtual image at a finite distance.

Teacher's Note:
1. Object inside the focus gives a virtual image.

Answer: (c) 15°

Explanation:
1. The ray enters undeviated and meets the second face at an angle of incidence equal to the apex angle, 30°.
2. \( \sin r = \sqrt{2} \sin 30^{\circ} = 0.707 \), so \( r = 45^{\circ} \).
3. Deviation = \( r \) – \( A = 45^{\circ} \) – \( 30^{\circ} = 15^{\circ} \).

Teacher's Note:
1. For normal incidence on the first face, the deviation is \( r \) – \( A \).

Answer: (a) 5R

Explanation:
1. Refraction at a single surface: \( \frac{1.5}{v} - \frac{1}{u} = \frac{1.5 - 1}{R} \).
2. With \( u \) = –d and \( v = +d \): \( 1.5/d + 1/d = 0.5/R \), so \( d = 5R \).

Teacher's Note:
1. Take the direction of the incident light as positive.

Answer: (a) L/l

Explanation:
1. The objective lens is at distance f_o + f_e from the eyepiece. The eyepiece magnification for this object is \( l/L \) = f_e/f_o.
2. Telescope magnification = f_o/f_e = \( L/l \).

Teacher's Note:
1. Treat the objective as an object for the eyepiece.

Answer: (a) tan i

Explanation:
1. The reflected ray makes angle \( i \) with the normal and the refracted ray makes angle \( r \), with \( i + 90^{\circ} + r = 180^{\circ} \), so \( r = 90^{\circ} \) – \( i \).
2. Snell's law: \( \mu = \sin i/\sin r = \sin i/\cos i = \tan i \).

Teacher's Note:
1. This is Brewster's law: \( \mu = \tan \) i_B.

Answer: (d) a large, real image of S will be formed behind the eyepiece, far away from it

Explanation:
1. Now the image from the objective lies just beyond the focus of the eyepiece.
2. The eyepiece then forms a real, highly magnified image far behind it, since \( 1/v \) = 1/f_e – 1/(f_e + δ) is small.

Teacher's Note:
1. Object just outside the focus gives a large real image far away.

Answer: (c) away from the lens by t(1 – 1/μ)

Explanation:
1. A slab in a converging beam shifts the image away from the lens along the axis by the normal shift t(1 – 1/μ).

Teacher's Note:
1. Image moves in the direction of the light.

Answer: (d) away from P by t(1 – 1/μ)

Explanation:
1. A slab of thickness \( t \) shifts the apparent position of an object by t(1 – 1/μ) towards the observer (here the mirror).
2. So the mirror sees \( P \) at a distance smaller by t(1 – 1/μ).
3. To restore the original object distance, the mirror must be moved away from \( P \) by t(1 – 1/μ).

Teacher's Note:
1. Normal shift through a slab = t(1 – 1/μ).

Answer: (c) R/(μ – 1)

Explanation:
1. Applying the refraction formula at the two surfaces in turn, and setting the final image distance equal to the object distance \( x \), gives x = R/(μ – 1).
2. Check for \( \mu = 1.5 \): \( x = 2R \), and the symmetric path works out consistently.

Teacher's Note:
1. Use \( \mu_{2}/v \) – μ₁/u = (μ₂ – μ₁)/R twice.

Answer: (b) π – 2c

Explanation:
1. For incidence \( i \) less than \( c \) the deviation is \( r \) – \( i \), which is largest, π/2 – \( c \), at \( i = c \).
2. For \( i \) greater than \( c \) the ray is totally reflected and the deviation is π – \( 2i \), which is largest, π – \( 2c \), at \( i = c \).
3. Since π – \( 2c \) is greater than π/2 – \( c \), the maximum possible deviation is π – \( 2c \).

Teacher's Note:
1. Compare deviations for refraction and total internal reflection.

Answer: (d) b\left(\frac{f}{u-f}\right)^2

Explanation:
1. Longitudinal magnification = dv/du. From \( 1/v + 1/u = 1/f \), dv/du = –v²/u² = –(f/(u – f))².
2. So the image length is b(f/(u – f))².

Teacher's Note:
1. Differentiate the mirror formula.

Answer: (d) –6.67

Explanation:
1. Powers: \( +1/0.40 = +2.5 D \), –1/0.40 = –2.5 \( D \), –1/0.15 = –6.67 \( D \).
2. Total power = 2.5 – 2.5 – 6.67 = –6.67 \( D \).

Teacher's Note:
1. Powers add for thin lenses in contact.

Answer: (d) The air bubble cannot form a real image

Explanation:
1. The bubble behaves like a sphere of relative index 3/4, which is less than 1, so it acts as a diverging system.
2. Solving as for a sphere gives only negative (virtual) object distances, so no real image can be formed.

Teacher's Note:
1. Air in water behaves as a diverging lens.