JEE Physics Ray Optics and Optical Instruments MCQs Set 11

Answer: (d) B and C only

Explanation:
1. The minimum mirror length is half the person's height: \( \frac{170}{2}=85\ \text{cm} \), so (B) is correct and (A) is wrong.
2. To see his feet the lower edge of the mirror must be at most halfway between the eye level and the ground: \( \frac{160}{2}=80\ \text{cm} \) or lower, so (C) is correct.
3. The top of the mirror must reach the midpoint of eye and top of head, \( \frac{160+170}{2}=165\ \text{cm} \), which agrees with a 85 cm mirror starting at 80 cm; a bottom at 85 cm would cut off his feet, so (D) is wrong.

Teacher's Note:
1. Mirror length needed is half the object's height, wherever he stands.
2. The mirror edge positions depend on the mid-points between the eye and the feet and between the eye and the top of the head.

Answer: (c) A, B and D only

Explanation:
1. The image S' of S lies 20 cm behind the wall at ground level. The man's eye at height h is 60 cm in front of the wall, so the line from his eye to S' is 80 cm long horizontally.
2. This line meets the wall at height \( y=h\times\frac{20}{80}=\frac h4 \).
3. The mirror extends from 40 cm to 80 cm, so \( 40\le\frac h4\le80 \) gives \( 160\le h\le320 \) cm.
4. So 160 cm, 280 cm and 320 cm are suitable, but 360 cm is not.

Teacher's Note:
1. The source's image behind a plane mirror is as far behind as the source is in front.
2. The ray must strike the mirror strip, so the point where the line of sight crosses the wall must lie between 40 cm and 80 cm.

Answer: (a) Virtual and erect, of magnification 1/2, and it moves towards the convex mirror as the object moves towards the plane mirror

Explanation:
1. The plane mirror forms the first image of A 10 cm behind it, which is \( 50+10=60\ \text{cm} \) from the convex mirror.
2. For the convex mirror, \( f=\frac{120}{2}=+60\ \text{cm} \) and \( u=-60\ \text{cm} \): \( \frac1v=\frac1f-\frac1u=\frac1{60}+\frac1{60}\Rightarrow v=30\ \text{cm} \), behind the mirror, so the image is virtual.
3. \( m=-\frac vu=\frac{30}{60}=\frac12 \), erect (the plane mirror does not change size).
4. As the object moves towards the plane mirror, its first image moves towards the convex mirror, so \( |u| \) decreases and \( v=\frac{fu}{f+u} \) (magnitudes) decreases: the final image moves towards the convex mirror.

Teacher's Note:
1. Treat the plane-mirror image as the object for the second mirror.
2. A convex mirror always gives a virtual, erect, diminished image of a real object, and the image shifts towards the mirror as the object moves closer.

Answer: (c) 9 cm/s away from the mirror

Explanation:
1. With \( u=-20,\ f=-12 \): \( \frac1v=\frac1f-\frac1u=-\frac1{12}+\frac1{20}=-\frac1{30} \), so the image is real at 30 cm.
2. Differentiating the mirror formula: \( v_I=-\left(\frac vu\right)^2 v_O=-\left(\frac{30}{20}\right)^2\times4 \) in magnitude \( 9\ \text{cm/s} \).
3. The object is between F and C, so as it approaches the mirror the real image moves away from it.

Teacher's Note:
1. The longitudinal velocity of the image is (v/u)^2 times the object's velocity.
2. For a real object in front of a concave mirror the object and the real image move in opposite directions (relative to the mirror).

Answer: (a) \( t\left(1-\frac1\mu\right) \), away from the plate

Explanation:
1. A plate of thickness t shifts the image of a real object towards the plate by \( t\left(1-\frac1\mu\right) \).
2. For a converging beam (a virtual object) the displacement is in the direction of the light, so the convergence point moves away from the plate by the same amount.

Teacher's Note:
1. The normal shift of a plate is always t(1 - 1/μ), in the direction of the incident light for the plate's effect on rays.
2. Real objects appear nearer; converging beams focus farther.

Answer: (a) B, C and D only

Explanation:
1. Let the angle of incidence at the 1-2 interface be θ. The two interfaces are perpendicular, so the angle of incidence at the 1-3 interface is \( 90°-\theta \).
2. TIR at both: \( \sin\theta>\frac{\mu_2}{\mu_1} \) and \( \cos\theta>\frac{\mu_3}{\mu_1} \).
3. Squaring and adding: \( 1>\frac{\mu_2^2+\mu_3^2}{\mu_1^2} \), i.e. \( \mu_1^2>\mu_2^2+\mu_3^2 \), which gives (B) and (C).
4. Since \( \mu_1>\mu_3 \), (D) also holds. (A) is not necessary because nothing fixes the order of \( \mu_2 \) and \( \mu_3 \).

Teacher's Note:
1. After reflection from a perpendicular interface the two angles of incidence are complementary.
2. Add sin^2 + cos^2 = 1 to combine the two TIR conditions.

Answer: (c) A and C only

Explanation:
1. At P the angle of incidence is 60°. The critical angle: \( \sin\theta_c=\frac{4/3}{5/3}=0.8 \) and \( \sin60°=0.866>0.8 \), so the ray is totally reflected at P.
2. The reflected ray strikes the base at 30° to the normal. Snell's law: \( \frac53\sin30°=\frac43\sin\theta_2\Rightarrow\sin\theta_2=\frac58 \) (A).
3. TIR at P stops when \( \frac{\mu_w}{5/3}\ge\sin60° \), i.e. \( \mu_w\ge\frac53\cdot\frac{\sqrt3}2=\frac{5}{2\sqrt3} \) (C). 5/6 is smaller than the present 4/3, so (D) is wrong.

Teacher's Note:
1. Find the angle of incidence on each face from the geometry before applying Snell's law.
2. TIR at a glass-liquid face stops when the liquid index reaches μ_glass × sin(angle of incidence).

Answer: (b) A, B and D only

Explanation:
1. The i-δ curve has a single minimum; for any larger δ there are two angles of incidence (i and e interchanged), so (A) is correct.
2. At minimum deviation the ray inside is symmetric about the prism, hence parallel to the base of an isosceles (equilateral) prism: (B) is correct.
3. Maximum deviation occurs only at grazing incidence, which is a single angle, so (C) is wrong.
4. From \( \mu=\frac{\sin\frac{A+\delta_m}2}{\sin\frac A2} \), a larger relative index gives a larger \( \delta_m \): (D) is correct.

Teacher's Note:
1. Because of the reversibility of light, the i-δ graph is symmetric about the minimum.
2. Minimum deviation: i = e and r1 = r2 = A/2.

Answer: (a) A and B only

Explanation:
1. First refraction: \( \frac{3/2}{v}-\frac{1}{-d_1}=\frac{3/2-1}{60} \), so \( \frac{3}{2v}=\frac1{120}-\frac1{d_1} \).
2. d1 = 120 cm: v = ∞, the rays are parallel in the glass, reflect from the plane face as a parallel beam and come back out to focus at 120 cm, which is O, whatever d2 is. (A) is correct.
3. d1 = 240 cm: \( \frac3{2v}=\frac1{240} \), v = 360 cm. The image I1 lies 360 cm from the pole. After the plane mirror the image is at \( 2d_2-360 \).
4. For the final image to coincide with O the reflected image must again be at 360 cm, so \( 2d_2-360=360\Rightarrow d_2=360\ \text{cm} \). (B) is correct; (C) and (D) are wrong.

Teacher's Note:
1. The light path is reversible: an object at 120 cm gives a parallel beam inside the glass.
2. A plane mirror at distance d2 maps an image at x to 2·d2 − x.

Answer: (c) A and C only

Explanation:
1. Taking light as travelling from right to left, the centre of curvature lies on the incident side, so R is negative: \( \frac{\mu_2}{v}=\frac{\mu_1}{u}+\frac{\mu_2-\mu_1}{R} \).
2. μ2 > μ1, real object (u < 0): both terms on the right are negative, so v < 0, a virtual image always. (A) is correct; for a virtual object (u > 0) the sum can be positive, so a real image is possible and (B) is wrong.
3. μ1 > μ2, virtual object (u > 0): \( \frac{\mu_1}{u}>0 \) and \( \frac{\mu_2-\mu_1}{R}>0 \) so v > 0 always, a real image only. (C) is correct.
4. For a real object with μ1 > μ2 the two terms have opposite signs, so a real image is possible and (D) is wrong.

Teacher's Note:
1. Decide the sign of R from the direction of the light and the side on which the centre of curvature lies.
2. For virtual or real objects, check the sign of each term in the refraction formula.

Answer: (a) a concave mirror of focal length 18 cm

Explanation:
1. Powers of the refracting surfaces (centre on the incident side taken negative): left surface \( \frac{1.5-1}{-10}=-0.05\ \text{cm}^{-1} \), right glass-water surface \( \frac{4/3-3/2}{+15}=-0.0111\ \text{cm}^{-1} \).
2. Mirror in water: \( \frac{2n}{R}=\frac{2\times4/3}{15}=+0.1778\ \text{cm}^{-1} \).
3. Light passes the two refracting surfaces twice and reflects once: \( P=2(-0.05-0.0111)+0.1778=0.0556=\frac1{18} \).
4. A positive net power means a converging system, that is a concave mirror of focal length 18 cm.

Teacher's Note:
1. For a system containing a mirror, the refracting surfaces are crossed twice, so their powers count twice.
2. Positive total power with reflection means the system acts as a concave mirror.

Answer: (b) A, B, C and D

Explanation:
1. The two magnifications are \( m \) and \( \frac1m \) with ratio 4, so \( m^2=4 \): the magnifications are 2 and 1/2. The object is 2 times the shorter image (A).
2. For m = 2: \( v=2u,\ u+v=96 \) gives \( u=32,\ v=64 \). The other position has \( u=64,\ v=32 \), so the lens moves by \( 64-32=32\ \text{cm} \) (B).
3. \( f=\frac{32\times64}{96}=\frac{64}{3}\ \text{cm} \) (C).
4. The shorter image (m = 1/2) is formed with v = 32 cm, the lens-to-screen distance (D).

Teacher's Note:
1. The two images have magnifications m and 1/m, so their ratio is m^2.
2. The shorter image corresponds to the lens position nearer the screen.

Answer: (a) B and C only

Explanation:
1. Lens (first pass): \( u=-20,\ f=10 \) gives \( v=20\ \text{cm} \), \( m_1=-1 \).
2. Mirror: object distance \( 40-20=20\ \text{cm} \), \( \frac1v=\frac1{7.5}-\frac1{20}=\frac1{12} \), \( v=12\ \text{cm} \), \( m_2=-\frac{12}{20}=-0.6 \).
3. The mirror image is 28 cm from the lens; lens (second pass): \( \frac1v=\frac1{10}-\frac1{28}=\frac{9}{140} \), \( v=15.56\ \text{cm} \) to the left, \( m_3=-\frac{15.56}{28}=-0.556 \).
4. Overall \( m=(-1)(-0.6)(-0.556)=-0.333 \); the image is real and inverted. So (B) and (C) are correct.

Teacher's Note:
1. Treat each pass through a lens or reflection as a separate step and multiply the magnifications.
2. The image formed by the mirror becomes the object for the lens in the return pass.

Answer: (d) A and C only

Explanation:
1. A is a half cut along the axis: both curved surfaces remain, so its focal length (and power) is unchanged, \( P_A=P \) (A).
2. B and C each have only one curved surface, so each has half the power: \( P_B=P_C=\frac P2 \) (C is correct; D is wrong).

Teacher's Note:
1. Cutting a lens along its axis does not change the focal length (only the brightness changes).
2. A plano-convex half has half the power of the equiconvex lens.

Answer: (d) f_o = 30 cm and f_e = 6 cm

Explanation:
1. For normal adjustment \( f_o+f_e=36\ \text{cm} \) and \( \frac{f_o}{f_e}=5 \).
2. So \( 6f_e=36 \), giving \( f_e=6\ \text{cm},\ f_o=30\ \text{cm} \).

Teacher's Note:
1. In an astronomical telescope both lenses are converging, so both focal lengths are positive.
2. Magnification is f_o/f_e and the tube length is f_o + f_e.

Answer: (b) A, B, C and D

Explanation:
1. A lens has a different refractive index for each colour, so each colour has its own focal length: this is chromatic aberration (A, C).
2. The separation of the red and violet foci is the longitudinal chromatic aberration, \( f_r-f_v=\omega f \) with ω the dispersive power (B).
3. A spectrometer produces and studies pure spectra (D).

Teacher's Note:
1. Chromatic aberration is a defect of lenses, not mirrors.
2. The measure of axial chromatic aberration is the product of dispersive power and mean focal length.

Answer: (b) \( \frac{tD}{d} \)

Explanation:
1. Extra optical path through the upper sheet: \( (2\mu-1)t \). Through the lower sheet: \( (\mu-1)(2t)=(2\mu-2)t \).
2. The upper path exceeds the lower by \( t \). The central maximum shifts to the side where the geometric path is shorter, that is towards the upper slit: \( y=\frac{tD}{d} \).

Teacher's Note:
1. The central maximum is where the total optical paths are equal.
2. It always shifts towards the slit that carries the extra optical path.

Answer: (a) B, C and D only

Explanation:
1. At the centre all wavelengths have zero path difference, so it is white.
2. Violet has the smallest fringe width, so its first maximum is nearest the centre. The colours overlap at different positions, so no point is dark for every colour.

Teacher's Note:
1. Fringe width is proportional to wavelength.
2. The shortest wavelength forms its first bright fringe closest to the centre.

Answer: (b) 81 : 1

Explanation:
1. Ray AB: \( I_1=0.2I \). Ray A'B': transmitted 0.8I, reflected 20% at the lower face gives 0.16I, transmitted again at the top gives \( I_2=0.8\times0.16I=0.128I \).
2. Amplitude ratio \( \frac{a_1}{a_2}=\sqrt{\frac{0.2}{0.128}}=1.25 \).
3. \( \frac{I_{max}}{I_{min}}=\left(\frac{a_1+a_2}{a_1-a_2}\right)^2=\left(\frac{2.25}{0.25}\right)^2=81 \).

Teacher's Note:
1. Intensity is proportional to the square of amplitude.
2. Track the fraction transmitted or reflected at each surface.

Answer: (a) A and C bright, B and D dark

Explanation:
1. On the perpendicular bisector the path difference is zero, so A and C are bright.
2. At B and D the path difference equals the separation, \( 5\ \mu m=2.5\lambda \), which is a half-integer number of wavelengths, so B and D are dark.

Teacher's Note:
1. Zero path difference gives a bright point.
2. A half-odd multiple of λ gives a dark point.

Answer: (d) 60

Explanation:
1. Separation \( d=3a \), so \( \frac d\lambda=15 \). Orders of maxima run from m = −15 to +15.
2. The orders m = ±15 occur only once each (along the axis), while the other 29 orders occur at two points each: \( 29\times2+2=60 \).

Teacher's Note:
1. The path difference on a far circle is d cosθ, which ranges from −d to +d.
2. The extreme orders touch the circle at one point only.

Answer: (b) I

Explanation:
1. Amplitudes are proportional to \( \sqrt I \) and \( 2\sqrt I \).
2. \( I_{res}=(2\sqrt I-\sqrt I)^2=I \) at phase difference π.

Teacher's Note:
1. Resultant amplitude at phase π is the difference of amplitudes.
2. Intensity is the square of amplitude.

Answer: (d) 0.225 mm

Explanation:
1. In water \( \lambda'=\frac{600}{4/3}=450\ \text{nm} \).
2. \( \beta=\frac{\lambda' D}{d}=\frac{450\times10^{-9}\times1}{2\times10^{-3}}=0.225\ \text{mm} \).

Teacher's Note:
1. Wavelength in a medium is λ divided by μ.
2. Use β = λD/d with the new wavelength.

Answer: (b) \( 8\ \mu m \)

Explanation:
1. The extra optical path is \( (1.7-1.4)t=5\lambda \).
2. \( t=\frac{5\times4800\times10^{-10}}{0.3}=8\times10^{-6}\ \text{m}=8\ \mu m \).

Teacher's Note:
1. The difference in optical path equals 5 wavelengths for the 5th bright fringe.
2. Convert Å to metres carefully.

Answer: (a) 0.2 mm

Explanation:
1. \( I=I_{max}\cos^2\frac\phi2=0.75I_{max} \) gives \( \frac\phi2=30^\circ \), so \( \phi=\frac\pi3 \) and path difference \( \frac\lambda6 \).
2. \( y=\frac{\lambda D}{6d}=0.1\ \text{mm} \) on each side of the centre, so the two points nearest each other, on opposite sides, are \( 0.2\ \text{mm} \) apart.

Teacher's Note:
1. 75% intensity corresponds to cos²(φ/2) = 3/4.
2. The two nearest points lie symmetrically about the central maximum.