JEE Physics Ray Optics and Optical Instruments MCQs Set 07

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

Explanation:
1. The ray makes \(30^\circ\) with the normal, so after reflection from M₁ it makes \(60^\circ\) with M₁ and moves towards M₂.
2. For the final ray to be parallel to M₁, using the rule that two reflections turn the ray through \(2\theta\) with respect to the first direction, we need \(2\theta + 60^\circ = 180^\circ\).
3. This gives \(\theta = 60^\circ\).

Teacher's Note:
1. Draw the reflected ray and apply the law of reflection again at M₂.
2. The angle between the ray and M₂ is the same before and after reflection.

Answer: (a) One

Explanation:
1. The two mirrors do not face each other along the full length, so the regions from which the images can be seen (fields of view) are separate and do not overlap.
2. At any position an observer receives light by only one route.
3. So at most one image of O can be seen at a time.

Teacher's Note:
1. Parallel mirrors give infinite images only when they overlap and face each other.
2. Here the mirrors are staggered, so the fields of view do not intersect.

Answer: (b) 1.78 cm

Explanation:
1. For the convex mirror \(f=+12.5\) cm and \(u=-40\) cm.
2. \(\dfrac1v=\dfrac1f-\dfrac1u=\dfrac1{12.5}+\dfrac1{40}=\dfrac{21}{200}\), so \(v=9.52\) cm.
3. \(m=-v/u=0.238\), so image height \(=7.5\times0.238=1.78\) cm.

Teacher's Note:
1. A convex mirror always gives a small, erect, virtual image.
2. Use the sign convention carefully with f positive.

Answer: (c) 10 mm

Explanation:
1. The square lies with side AD along the axis. The near side CD (perpendicular to the axis) is at \(u=-15\) cm: \(v=-30\) cm, \(m=-2\), so its image is 2 mm.
2. The far side AB is at \(u=-15.1\) cm: \(v=-29.61\) cm, \(m\approx-1.96\), so its image is about 1.96 mm.
3. The two sides along the axis have images of length \(30-29.61=0.39\) cm \(\approx3.9\) mm each (slightly tilted).
4. Perimeter \(\approx 2+1.96+3.9+3.9\approx11.8\) mm, so the nearest option is 10 mm.

Teacher's Note:
1. Transverse and axial sides of an object are magnified differently by a mirror.
2. Treat each side separately and add the image lengths.

Answer: (a) 3d

Explanation:
1. The image S' is at distance L behind the mirror.
2. The man can see the image only when the line from his eye to S' passes through the mirror.
3. The distance of the man from S' is \(3L\), so by similar triangles the length of his path is \(d\times\dfrac{3L}{L}=3d\).

Teacher's Note:
1. Draw the image and join it to the edges of the mirror.
2. Use similar triangles to find the length at the man's distance.

Answer: (b) \(\sin x=\dfrac58\sin\theta\)

Explanation:
1. For a stack of parallel layers, Snell's law gives \(\mu_1\sin\theta_1=\mu_2\sin\theta_2=\ldots\), so only the first and last media matter.
2. \(\mu\sin\theta = 1.6\mu\sin x\).
3. \(\sin x=\dfrac{\sin\theta}{1.6}=\dfrac58\sin\theta\).

Teacher's Note:
1. The product \(\mu\sin\theta\) stays constant across parallel interfaces.
2. The intermediate layers do not affect the final angle.

Answer: (d) \(\dfrac\pi2-C\)

Explanation:
1. The deviation of the refracted ray is \(\delta = r-i\).
2. Refraction exists only for \(i\le C\). The deviation is largest at \(i=C\), where \(r=90^\circ\).
3. So \(\delta_{max}=\dfrac\pi2-C\).

Teacher's Note:
1. Deviation grows as the incidence angle rises towards the critical angle.
2. At exactly C the refracted ray grazes the surface.

Answer: (b) \(178^\circ\) clockwise

Explanation:
1. The prism deviation is \(\delta_p=(\mu-1)A=0.5\times4^\circ=2^\circ\) clockwise (towards the base).
2. The ray then meets the vertical mirror at an angle of incidence \(i=2^\circ\), so the mirror deviates it by \(180^\circ-2i=176^\circ\), also clockwise.
3. Total deviation \(=2^\circ+176^\circ=178^\circ\) clockwise.

Teacher's Note:
1. A thin prism deviates a ray towards its base.
2. A mirror turns a ray through 180° minus twice the angle of incidence.

Answer: (c) image is always virtual

Explanation:
1. Taking the light to travel from X to Y with \(R=-10\) cm and \(u=-x\): \(\dfrac{3/2}{v}-\dfrac{4/3}{-x}=\dfrac{3/2-4/3}{-10}\).
2. This gives \(\dfrac{3}{2v}=-\dfrac1{60}-\dfrac4{3x}\), which is negative for every positive x.
3. So v is always negative, i.e. the image is always virtual.

Teacher's Note:
1. If the surface is concave towards the object medium and the second medium has the larger index, the image is virtual for every real object.
2. Check the sign of v for all values of x.

Answer: (d) a convergent and a divergent beam

Explanation:
1. The upper half of the lens is surrounded by liquid of index \(\mu_2>\mu_1\), so it behaves like a diverging lens.
2. The lower half is surrounded by liquid of index \(\mu_3<\mu_1\), so it behaves like a converging lens.
3. So one divergent and one convergent beam are produced.

Teacher's Note:
1. A convex lens behaves as a concave lens in a denser medium.
2. The two halves act separately.

Answer: (c) f₁ + f₂

Explanation:
1. A ray parallel to the axis emerges parallel to the axis after both lenses.
2. This happens only if the second focus of L₁ and the first focus of L₂ coincide.
3. Hence the separation is \(f_1+f_2\).

Teacher's Note:
1. This arrangement is a telescope in normal adjustment.
2. The lens separation equals the sum of focal lengths.

Answer: (a) 20 cm

Explanation:
1. \(f=\dfrac1P=\dfrac15\) m \(=20\) cm.
2. For an equiconvex lens \(\dfrac1f=(\mu-1)\dfrac2R\), so \(\dfrac1{20}=0.5\times\dfrac2R\).
3. \(R=20\) cm.

Teacher's Note:
1. Power in dioptre is 1 divided by focal length in metre.
2. For an equiconvex lens \(R_1=R,\ R_2=-R\).

Answer: (d) \(\tan^{-1}(\sin\theta_C)\)

Explanation:
1. The reflected and refracted rays are at right angles, so the angle of refraction is \(r=90^\circ-A\).
2. Snell's law: \(\dfrac{\sin A}{\sin r}=\dfrac{\sin A}{\cos A}=\tan A=\dfrac{\mu_{rarer}}{\mu_{denser}}=\sin\theta_C\).
3. So \(A=\tan^{-1}(\sin\theta_C)\).

Teacher's Note:
1. This is the Brewster angle condition seen from the denser medium.
2. Here \(\tan A=\sin\theta_C\).

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

Explanation:
1. For total internal reflection the angle of incidence \(45^\circ\) must exceed the critical angle.
2. \(\sin45^\circ>\sin\theta_c=\dfrac1n\), so \(\dfrac1{\sqrt2}>\dfrac1n\).
3. Hence \(n>\sqrt2\).

Teacher's Note:
1. A 45-45-90 prism turns a ray through 90° only if n is greater than √2.
2. The critical angle of glass is about 42°.

Answer: (d) 70 cm from point B at right; real

Explanation:
1. For the convex lens \(u=-20\) cm, \(f=+5\) cm: \(v=\dfrac{uf}{u+f}=\dfrac{-100}{-15}=\dfrac{20}3\) cm to the right of A.
2. This image is \(\dfrac{20}{3}-2=\dfrac{14}3\) cm beyond the concave lens, so it is a virtual object for the second lens: \(u=+\dfrac{14}{3}\) cm, \(f=-5\) cm.
3. \(\dfrac1v=\dfrac1f+\dfrac1u=-\dfrac15+\dfrac3{14}=\dfrac1{70}\), so \(v=+70\) cm, a real image to the right of B.

Teacher's Note:
1. The image of the first lens acts as the object for the second.
2. An image formed beyond the second lens is a virtual object.

Answer: (a) +3.0 D

Explanation:
1. The lens must form a virtual image of the book at his near point, 100 cm.
2. \(u=-25\) cm, \(v=-100\) cm: \(\dfrac1f=\dfrac1v-\dfrac1u=-\dfrac1{100}+\dfrac1{25}=\dfrac3{100}\).
3. \(f=\dfrac{100}3\) cm, so \(P=\dfrac{100}{f}=+3.0\) D.

Teacher's Note:
1. A far-sighted eye needs a convex lens.
2. Image at the near point, object at 25 cm.

Answer: (a) 1.8 cm

Explanation:
1. For a relaxed eye the final image is at infinity, so the tube length is \(L=v_0+f_e\), giving \(v_0=14-5=9\) cm.
2. \(m=\dfrac{v_0}{u_0}\cdot\dfrac{D}{f_e}\): \(25=\dfrac9{u_0}\times\dfrac{25}5\).
3. \(u_0=\dfrac95=1.8\) cm.

Teacher's Note:
1. In normal adjustment the tube length equals \(v_0+f_e\).
2. Take D = 25 cm for the eye.

Answer: (a) 2.5

Explanation:
1. In a Galilean telescope in normal adjustment, \(L=f_o-f_e\), so \(f_e=15-9=6\) cm.
2. \(m=\dfrac{f_o}{f_e}=\dfrac{15}6=2.5\).

Teacher's Note:
1. A Galilean telescope uses a concave eyepiece, so the lenses are closer than \(f_o\).
2. It gives an erect image.

Answer: (a) 100 cm

Explanation:
1. For normal setting \(L=f_o+f_e=105\) cm and \(m=\dfrac{f_o}{f_e}=20\).
2. So \(f_o=20f_e\), \(21f_e=105\), \(f_e=5\) cm.
3. \(f_o=100\) cm.

Teacher's Note:
1. Two equations in two unknowns.
2. The objective has the much larger focal length.

Answer: (b) 36 cm away from the mirror

Explanation:
1. For \(u=-40\) cm: \(\dfrac1v=\dfrac1f-\dfrac1u\) gives \(\dfrac1v=-\dfrac1{15}+\dfrac1{40}\), so \(v=-24\) cm.
2. For \(u=-20\) cm: \(\dfrac1v=-\dfrac1{15}+\dfrac1{20}=-\dfrac1{60}\), so \(v=-60\) cm.
3. The image moves from 24 cm to 60 cm in front of the mirror, i.e. 36 cm away from the mirror.

Teacher's Note:
1. When the object moves closer to the focus the image moves away quickly.
2. Find both image positions and subtract.