JEE Physics Ray Optics and Optical Instruments MCQs Set 13

Answer: (c) A and B only

Explanation:
1. The path difference can reach at most d. For \( d=\lambda \) the first order lies at 90° (not observable), so only the central maximum is seen (A true). For \( \lambda2. Let slit 2 have intensity I2 and slit 1 have 4I2. Then \( I_{max}=9I_2 \) and \( I_{min}=I_2 \).
3. In (C) the slits become equal, so \( I_{max}=4I_2 \) and \( I_{min}=0 \): both decrease (C false). In (D) with slit 2 raised to 4I2, \( I_{max}=16I_2 \) rises but \( I_{min}=0 \) falls (D false).

Teacher's Note:
1. Order n needs nλ ≤ d.
2. Compare (√I1 ± √I2)² before and after.

Answer: (a) I(P2) > I(P1)

Explanation:
1. The sheet adds an optical path λ/2 to S2. The net path difference (S2 side minus S1 side) is \( \frac\lambda2-\Delta x \) where \( \Delta x=S_1P-S_2P \).
2. At P0: \( \frac\lambda2 \), phase π, so intensity is zero. At P1: \( \frac\lambda2-\frac\lambda4=\frac\lambda4 \), phase π/2, so \( I=2I_0 \). At P2: \( \frac\lambda2-\frac\lambda3=\frac\lambda6 \), phase π/3, so \( I=3I_0 \).
3. Thus \( I(P_2)>I(P_1) \); P0 is dark, δ(P0) is not zero, and I(P1) is not zero.

Teacher's Note:
1. The sheet shifts the whole pattern by half a fringe.
2. I = 4I0 cos²(φ/2) for equal slits.

Answer: (c) \( \beta_R>\beta_G>\beta_B \)

Explanation:
1. Fringe width \( \beta=\frac{\lambda D}{d}\propto\lambda \).
2. Wavelengths: red > green > blue, so \( \beta_R>\beta_G>\beta_B \).

Teacher's Note:
1. Fringe width is proportional to wavelength.
2. Red has the longest wavelength.

Answer: (a) decreases

Explanation:
1. \( d=\frac{\lambda}{2\sin\theta} \), so \( \frac{\Delta d}{d}=\cot\theta\ \Delta\theta \).
2. Since cotθ decreases as θ increases from 0° to 90°, the fractional error decreases.

Teacher's Note:
1. Differentiate ln d = ln(λ/2) − ln sinθ.
2. cotθ decreases with θ.

Answer: (d) B and D only

Explanation:
1. With θ measured from the bisector, bright spots obey \( d\sin\theta=m\lambda \), so \( \Delta\theta=\frac{\lambda}{d\cos\theta} \), which increases towards P2 (A false).
2. At P2, \( \theta=90^\circ \) and \( m=\frac d\lambda=\frac{1.8\times10^{-3}}{600\times10^{-9}}=3000 \), an integer, so P2 is a bright spot of maximum order (B true, C false).
3. From P1 (m = 0) to P2 (m = 3000) there are about 3000 fringes (D true).

Teacher's Note:
1. Path difference is d sinθ from the bisector.
2. d/λ = 3000 is an integer, so P2 is bright.

Answer: (d) 5 m

Explanation:
1. Limit of resolution of the eye: \( \theta=\frac{1.22\lambda}{D}=\frac{1.22\times500\times10^{-9}}{3\times10^{-3}}=2.0\times10^{-4}\ \text{rad} \).
2. The dots are resolved if \( \frac{1\ \text{mm}}{L}\ge\theta \), so \( L_{max}=\frac{10^{-3}}{2.03\times10^{-4}}\approx5\ \text{m} \).

Teacher's Note:
1. Use the Rayleigh criterion θ = 1.22 λ/D.
2. The angle subtended by the dots is separation divided by distance.

Answer: (b) \( n>\sqrt2 \)

Explanation:
1. The ray hits the hypotenuse at 45°. TIR requires \( 45°>\theta_c \), that is \( \sin45°>\frac1n \), so \( n>\sqrt2 \).

Teacher's Note:
1. TIR needs the angle of incidence to exceed the critical angle.
2. sin θc = 1/n.

Answer: (d) −10 cm

Explanation:
1. Total power \( P=-15+5=-10\ \text{D} \).
2. Focal length \( f=\frac1P=-0.1\ \text{m}=-10\ \text{cm} \).

Teacher's Note:
1. Powers add for lenses in contact.
2. A negative power gives a diverging combination.

Answer: (b) \( \sin^{-1}\left(\frac1{\sqrt3}\right) \)

Explanation:
1. Grazing along the wall means the ray meets the wall at the critical angle: \( \sin C=\frac1n=\frac{\sqrt3}2 \), so \( C=60° \) and the refraction angle at the end is \( r=90°-60°=30° \).
2. At the end face: \( \sin\theta=n\sin r=\frac2{\sqrt3}\times\frac12=\frac1{\sqrt3} \).

Teacher's Note:
1. The angle of incidence on the side wall is the complement of the refraction angle at the end face.
2. Grazing along the wall corresponds to the critical angle.

Answer: (d) increase

Explanation:
1. Red light has a smaller refractive index than blue in glass.
2. Since \( \frac1f=(\mu-1)\left(\frac1{R_1}-\frac1{R_2}\right) \), a smaller μ gives a larger f: the focal length increases.

Teacher's Note:
1. Focal length varies inversely with (μ − 1).
2. This is the cause of chromatic aberration.

Answer: (c) 45°

Explanation:
1. \( |\vec A|=\sqrt{108+192+100}=20 \). The normal to the boundary is along z, so \( \cos i=\frac{10}{20} \) and \( i=60° \).
2. Snell's law: \( \sqrt2\sin60°=\sqrt3\sin r \), so \( \sin r=\frac{\sqrt2\times\sqrt3/2}{\sqrt3}=\frac1{\sqrt2} \) and \( r=45° \).

Teacher's Note:
1. The angle of incidence is the angle between the ray and the normal to the boundary.
2. The component of the vector along the normal gives cos i.

Answer: (b) \( \frac1{15}\ \text{m/s} \)

Explanation:
1. Convex mirror: \( f=+0.2\ \text{m},\ u=-2.8\ \text{m} \) so \( \frac1v=\frac1{0.2}+\frac1{2.8}=\frac{15}{2.8} \) and \( v=0.187\ \text{m} \).
2. The magnification factor is \( \frac vu=\frac{f}{f-u}=\frac{0.2}{3.0}=\frac1{15} \), so the image speed is \( \left(\frac1{15}\right)^2\times15=\frac1{15}\ \text{m/s} \).

Teacher's Note:
1. For a mirror, the image speed is (v/u)² times the object speed, and v/u = f/(f − u) for the mirror formula.
2. The image of a far object barely moves in a convex mirror.

Answer: (d) f1 > f and f2 becomes negative

Explanation:
1. Lens in a liquid: \( \frac1{f_l}=\left(\frac{\mu_g}{\mu_l}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right) \) and in air \( \frac1f=(\mu_g-1)(\ldots) \).
2. Liquid 4/3: \( \frac f{f_1}=\frac{1.5/(4/3)-1}{0.5}=0.25 \), so \( f_1=4f>f \).
3. Liquid 5/3: \( \frac f{f_2}=\frac{1.5/(5/3)-1}{0.5}=-0.2 \), so \( f_2=-5f \) (negative).

Teacher's Note:
1. If the liquid is denser than the lens material, a convex lens behaves as a diverging lens.
2. Compare the ratio μg/μl with 1.

Answer: (b) bends upwards

Explanation:
1. Light travels faster where the refractive index is smaller. The lower part of each wavefront (near the ground) therefore moves faster than the upper part.
2. The wavefront tilts so that its normal, that is the ray, turns towards the region of higher refractive index, upwards.

Teacher's Note:
1. Rays bend towards the region of larger refractive index (slower speed).
2. This is the opposite of the mirage seen on hot roads for rays coming down from the sky.

Answer: (d) 20 times nearer

Explanation:
1. A telescope magnifies the angle subtended by a distant object by the magnifying power, so the tree subtends 20 times the angle.
2. Equal angle at 20 times smaller distance would look the same, so the tree appears 20 times nearer (its real height is unchanged).

Teacher's Note:
1. An angular magnification of 20 is equivalent to viewing the tree from one-twentieth of the distance.
2. Telescopes do not change the actual size of the object.

Answer: (c) 1.5

Explanation:
1. Angle of the prism \( A=i+e-\delta=35°+79°-40°=74° \).
2. The minimum deviation cannot exceed the observed 40°, so \( \mu=\frac{\sin\frac{A+\delta_m}2}{\sin\frac A2}\le\frac{\sin57°}{\sin37°}=\frac{0.839}{0.602}=1.39 \).
3. Among the options the value closest to this upper limit is 1.5.

Teacher's Note:
1. The measured deviation is at least the minimum deviation.
2. Compute the maximum value of the formula μ = sin((A+δm)/2)/sin(A/2) with δm replaced by 40°.

Answer: (a) 10 cm and −15 cm

Explanation:
1. Let the concave power be \( P \) in magnitude; the convex power is \( \frac32P \).
2. Combination: \( \frac32P-P=\frac P2=\frac1{30} \), so \( P=\frac1{15} \) and the convex power is \( \frac1{10} \).
3. So \( f_1=10\ \text{cm},\ f_2=-15\ \text{cm} \).

Teacher's Note:
1. Powers of lenses in contact add, taking the concave power as negative.
2. The combination is converging, so the convex lens is the stronger one.

Answer: (c) 12 cm left of the lens

Explanation:
1. A plano-convex lens silvered on the plane side acts as a mirror of power \( P=\frac2f=\frac2{15} \), so \( f_{eq}=7.5\ \text{cm} \).
2. Mirror formula (magnitudes): \( \frac1v=\frac1{7.5}-\frac1{20}=\frac1{12} \), so \( v=12\ \text{cm} \) in front of the lens.

Teacher's Note:
1. The plane mirror adds no power; the lens is crossed twice.
2. The image is real and in front of the system.

Answer: (a) a reflected ray and a refracted ray, with the angle between them less than 180° − 2θ

Explanation:
1. Below the critical angle both partial reflection and refraction occur.
2. The reflected ray makes angle θ with the normal and the refracted ray makes angle r > θ (water to air). The angle between them is \( 180°-(\theta+r)<180°-2\theta \).

Teacher's Note:
1. Going from denser to rarer medium the refracted ray bends away from the normal.
2. The angle between the rays equals 180° minus (θ + r).

Answer: (d) f < x < 2f

Explanation:
1. When the eye moves to the left, the object nearer to the eye appears shifted to the right relative to the farther one. So the image is nearer to the eye (farther from the mirror) than the pin: \( v>x \).
2. For a real image formed farther from the mirror than the object, the object lies between f and 2f.

Teacher's Note:
1. Parallax: nearer objects appear to shift opposite to the movement of the eye.
2. If x = 2f the image coincides with the pin and no parallax is seen.

Answer: (c) 30° for both the colours

Explanation:
1. At minimum deviation the ray passes symmetrically, so \( r_1=r_2=\frac A2=30° \) for every colour.
2. The angles of incidence and deviation differ with colour, but the angle of refraction does not.

Teacher's Note:
1. At minimum deviation r1 = r2 = A/2.
2. Only the angle of incidence (and deviation) changes with the refractive index.

Answer: (b) \( \sin^{-1}\left(\frac18\right) \)

Explanation:
1. For parallel layers \( n\sin\theta \) is constant along the ray: \( n_0\sin\theta=n_{IV}\sin\theta_{IV} \).
2. The beam just misses entering Region IV when \( \theta_{IV}=90° \): \( n_0\sin\theta=\frac{n_0}8 \), so \( \sin\theta=\frac18 \).

Teacher's Note:
1. The beam is limited only by the last layer's refractive index.
2. The widths of the layers do not matter.

Answer: (a) C and D only

Explanation:
1. For f = 24 cm: \( v=\frac{uf}{u-f} \).
2. (42): \( v=\frac{42\times24}{18}=56 \) ok. (48): \( v=48 \) ok. (60): \( v=40 \) ok.
3. (66): \( v=\frac{66\times24}{42}=37.7 \), not 33 (error 4.7 cm). (78): \( v=\frac{78\times24}{54}=34.7 \), not 39 (error 4.3 cm).
4. Both exceed the permitted 0.2 cm error, so (66, 33) and (78, 39) are incorrect.

Teacher's Note:
1. Calculate v from the mirror formula for each u and compare with the recorded value.
2. The permitted error is only 0.2 cm.

Answer: (b) 6

Explanation:
1. Magnification of a real image: \( |m|=\frac{f}{|u|-f} \).
2. At 25 cm: \( \frac{20}{25-20}=4 \); at 50 cm: \( \frac{20}{50-20}=\frac23 \).
3. Ratio \( \frac{4}{2/3}=6 \).

Teacher's Note:
1. Use m = f/(u − f) with magnitudes.
2. The image shrinks as the object moves away.