JEE Physics Ray Optics and Optical Instruments MCQs Set 26

Answer: (b) 0.32 m

Explanation:
1. For an upright (virtual) image \( v=+5d \) behind the mirror and \( u=-d \), with \( f=-0.4 \) m.
2. \( \dfrac1{5d}-\dfrac1d=-\dfrac1{0.4}\Rightarrow -\dfrac4{5d}=-2.5\Rightarrow d=0.32 \) m.

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (d) \( 60^\circ \)

Explanation:
1. The incident ray is parallel to M2, so it meets M1 at an angle of incidence equal to the angle between the mirrors, \( \theta \) measured from the mirror surface; after reflection it makes angle \( \theta \) with M1 and meets M2.
2. The ray reflected from M2 is parallel to M1 only when the triangle formed by the two mirrors and the ray path is equilateral (each reflection turns the ray by \( 2\theta \) and the total turn must be \( 180^\circ-\theta \) with M1), so \( 3\theta=180^\circ \) and \( \theta=60^\circ \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (a) \( 30^\circ \)

Explanation:
1. Only the y-component reverses, so the mirror normal is along the y-axis.
2. \( \cos i=\hat n\cdot\hat d=\dfrac{\sqrt3}2\Rightarrow i=30^\circ \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (c) \( b\left(\dfrac{f}{u-f}\right)^2 \)

Explanation:
1. Differentiate the mirror formula: \( \dfrac{dv}{v^2}=-\dfrac{du}{u^2} \), so \( |dv|=\dfrac{v^2}{u^2}|du| \).
2. With \( v=\dfrac{uf}{u-f} \), \( \dfrac vu=\dfrac f{u-f} \), hence the image length is \( b\left(\dfrac f{u-f}\right)^2 \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (d) (66, 33) and (78, 39) only

Explanation:
1. Check \( \dfrac1u+\dfrac1v \) for each set. (42, 56): \( \dfrac1{24} \); (48, 48): \( \dfrac1{24} \); (60, 40): \( \dfrac1{24} \). These give \( f=24 \) cm.
2. (66, 33) gives \( \dfrac1{22} \), i.e. \( f=22 \) cm, and (78, 39) gives \( f=26 \) cm. Both differ from 24 cm by far more than the 0.2 cm error, so they cannot come from the experiment.

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (b) 3 km h\(^{-1}\)

Explanation:
1. \( f=+10 \) m. From \( \dfrac1u=\dfrac1f-\dfrac1v \): \( v=\dfrac{25}3 \) gives \( \dfrac1u=\dfrac1{10}-\dfrac3{25}=-\dfrac1{50} \), \( u=-50 \) m; \( v=\dfrac{50}7 \) gives \( \dfrac1u=\dfrac1{10}-\dfrac7{50}=-\dfrac1{25} \), \( u=-25 \) m.
2. The object moved 25 m in 30 s, i.e. \( \dfrac{25}{30}=0.833 \) m/s \( =3 \) km/h.

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (d) \( 2a+2b \)

Explanation:
1. Refractive index \( \mu=\dfrac{\sin60^\circ}{\sin30^\circ}=\sqrt3 \). Geometric path in vacuum \( =\dfrac a{\cos60^\circ}=2a \); in glass \( =\dfrac b{\cos30^\circ}=\dfrac{2b}{\sqrt3} \).
2. Optical path \( =2a+\sqrt3\cdot\dfrac{2b}{\sqrt3}=2a+2b \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (a) 1.36

Explanation:
1. Light is cut off at the critical angle C, so the spot radius is \( h\tan C=5.77 \) mm, giving \( \tan C=0.577 \), \( C=30^\circ \).
2. \( \mu_{liquid}=\mu_{block}\sin C=2.72\times0.5=1.36 \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (c) a reflected ray and a refracted ray, and the angle between them will be less than \( 180^\circ-2\theta \)

Explanation:
1. Below the critical angle both partial reflection and refraction occur.
2. The refracted angle satisfies \( r\gt \theta \) (water to air), so the angle between the rays is \( 180^\circ-\theta-r\lt 180^\circ-2\theta \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (d) 6 cm

Explanation:
1. Every ray from the centre meets the surface along the normal, so it passes undeviated.
2. The image therefore coincides with the object, at the centre, which is 6 cm from the surface.

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (b) \( \sqrt{\dfrac52} \)

Explanation:
1. Take the rod at the origin base, pin-hole at (2h, 3h). Empty beaker: the line to the rod top (0, h) has slope 1, so it meets the liquid level \( y=2h \) at \( x=h \).
2. In air the ray from (h, 2h) to the eye makes \( 45^\circ \) with the vertical; in liquid it goes from (h, 2h) to the rod foot (0, 0): \( \sin r=\dfrac h{\sqrt{h^2+4h^2}}=\dfrac1{\sqrt5} \). So \( \mu=\dfrac{\sin45^\circ}{\sin r}=\sqrt{\dfrac52} \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (c) \( \mu_4=\mu_1 \)

Explanation:
1. For parallel interfaces \( \mu_1\sin\theta_1=\mu_2\sin\theta_2=\mu_3\sin\theta_3=\mu_4\sin\theta_4 \).
2. The final ray is parallel to the first only if \( \theta_4=\theta_1 \), which needs \( \mu_4=\mu_1 \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (a) \( \alpha \)

Explanation:
1. A plane parallel slab does not change the direction of any ray; each ray emerges parallel to its incident direction (only displaced sideways).
2. Hence the divergence angle stays \( \alpha \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (c) \( \sin^{-1}(\tan r) \)

Explanation:
1. Reflected and refracted rays at \( 90^\circ \) give \( r+r'=90^\circ \), and \( r=i \). Snell: \( \mu\sin i=\sin r'=\cos i \), so \( \mu=\dfrac1{\tan i} \).
2. \( \sin C=\dfrac1\mu=\tan i=\tan r \), so \( C=\sin^{-1}(\tan r) \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (d) \( |f_1|=3R \) and \( |f_2|=2R \)

Explanation:
1. A thin film does not matter: the surface acts as an air-glass surface. Air to glass: \( \dfrac{1.5}{f_1}=\dfrac{0.5}R\Rightarrow f_1=3R \).
2. Glass to air the same surface gives \( \dfrac{1}{v}-\dfrac{1.5}{\infty}=\dfrac{1-1.5}{-R}\Rightarrow \dfrac1{f_2}=\dfrac{0.5}R \) (magnitude), so \( |f_2|=2R \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (b) 30 cm inside the slab (on the same side as O), virtual

Explanation:
1. Take light travelling from the slab to air: \( \dfrac{1}{v}-\dfrac{2}{u}=\dfrac{1-2}{R} \) with \( u=-15 \), and R = -10 cm (centre of curvature on the incident side).
2. \( \dfrac1v=-\dfrac{2}{15}+\dfrac1{10}=-\dfrac1{30}\Rightarrow v=-30 \) cm: a virtual image 30 cm from P on the object's side.

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (c) 4000 \(\mathrm{\AA}\) and \( 5\times10^{14} \) Hz

Explanation:
1. \( \lambda'=\dfrac\lambda\mu=\dfrac{6000}{1.5}=4000\ \mathrm{\AA} \).
2. Frequency is unchanged: \( f=\dfrac{3\times10^8}{6\times10^{-7}}=5\times10^{14} \) Hz.

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (a) \( \sqrt2 \)

Explanation:
1. At the end face \( \sin\alpha=n\sin r \), and the angle of incidence on the curved side is \( 90^\circ-r \). Total reflection needs \( n\cos r\ge1 \).
2. The worst case is \( \alpha=90^\circ \): \( \sin r=\dfrac1n \), so \( n\sqrt{1-\dfrac1{n^2}}=\sqrt{n^2-1}\ge1\Rightarrow n\ge\sqrt2 \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (b) \( \dfrac1{5\sqrt2}(3\hat i+4\hat j-5\hat k) \)

Explanation:
1. \( |\vec A|=\sqrt{108+192+100}=20 \), so \( \cos i=\dfrac{10}{20} \), \( i=60^\circ \). Snell: \( \sqrt2\sin60^\circ=\sqrt3\sin r\Rightarrow\sin r=\dfrac1{\sqrt2} \), \( r=45^\circ \).
2. The tangential part keeps its direction \( \dfrac{3\hat i+4\hat j}5 \): refracted unit vector \( =\sin45^\circ\dfrac{3\hat i+4\hat j}5-\cos45^\circ\hat k=\dfrac1{5\sqrt2}(3\hat i+4\hat j-5\hat k) \).

Teacher's Note:
1. For reflection and refraction at spherical surfaces use the sign convention consistently (distances measured from the pole, positive along the incident light) and apply the single-surface or mirror formula step by step.
2. Check the result with a quick ray argument: where a real image or a virtual image is expected, and whether the magnification is greater or less than one.

Answer: (d) \( \dfrac R{\mu_1-\mu_2} \)

Explanation:
1. \( \dfrac1f=\dfrac{\mu_1-1}R-\dfrac{\mu_2-1}R \) (the concave lens has negative power).
2. \( \dfrac1f=\dfrac{\mu_1-\mu_2}R\Rightarrow f=\dfrac R{\mu_1-\mu_2} \).

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (b) 3 : 1

Explanation:
1. Real image (m = -2): \( m=\dfrac f{f+u}\Rightarrow -2=\dfrac{20}{20+u}\Rightarrow u=-30 \) cm. Virtual image (m = +2): \( 2=\dfrac{20}{20+u}\Rightarrow u=-10 \) cm.
2. \( x_1:x_2=30:10=3:1 \).

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (c) \( \dfrac R{\mu_2-\mu_1} \)

Explanation:
1. The two curved surfaces coincide, so the combination is a single lens with powers \( \dfrac{\mu_2-1}R \) and \( -\dfrac{\mu_1-1}R \).
2. \( \dfrac1f=\dfrac{\mu_2-\mu_1}R\Rightarrow f=\dfrac R{\mu_2-\mu_1} \).

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (a) \( 1.16\times10^{-3} \) m/s towards the lens

Explanation:
1. \( v=\dfrac{uf}{u-f}\Rightarrow\dfrac{dv}{dt}=-\left(\dfrac f{u-f}\right)^2\dfrac{du}{dt} \) in magnitudes.
2. \( \left(\dfrac{0.3}{19.7}\right)^2\times5=1.16\times10^{-3} \) m/s; as the object recedes the image moves towards the lens.

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (b) \( 2\mu_1-\mu_2=1 \)

Explanation:
1. Magnitudes: \( f_1=\dfrac R{\mu_1-1} \), \( f_2=\dfrac R{\mu_2-1} \).
2. \( f_1=2f_2\Rightarrow\mu_2-1=2(\mu_1-1)\Rightarrow 2\mu_1-\mu_2=1 \).

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (d) real and at a distance of 40 cm from the convergent lens

Explanation:
1. The diverging lens forms a virtual image at 25 cm on the source side, i.e. \( 25+15=40 \) cm before the converging lens.
2. \( \dfrac1v=\dfrac1{20}-\dfrac1{40}=\dfrac1{40}\Rightarrow v=+40 \) cm: real, 40 cm beyond the convergent lens.

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (a) \( f_1\gt f \) and \( f_2 \) becomes negative

Explanation:
1. \( \dfrac1f\propto\dfrac{\mu}{\mu_m}-1 \): in air \( \tfrac12 \); in the liquid of 4/3: \( \tfrac98-1=\tfrac18 \) (smaller, positive).
2. So \( f_1=4f\gt f \). In the liquid of 5/3: \( \tfrac9{10}-1=-\tfrac1{10} \), negative, so \( f_2 \) is negative (diverging).

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.