JEE Physics Ray Optics and Optical Instruments MCQs Set 08

Answer: (d) 60 cm

Explanation:
1. The small displacement only shifts the images formed by the two halves sideways, it does not change the image distance.
2. Use the lens formula \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \) with \( u = -30 \) cm and \( f = 20 \) cm.
3. \( \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} \), so \( v = 60 \) cm.

Teacher's Note:
1. Sideways displacement changes the lateral position of the image, not the distance from the lens.
2. Billet's split lens uses this idea to make two coherent images.

Answer: (b) 21.4 cm

Explanation:
1. For the displacement method \( f = \frac{D^2 - d^2}{4D} \), where D is the object-screen distance and d the separation of the two lens positions.
2. \( f = \frac{90^2 - 20^2}{4\times90} = \frac{8100 - 400}{360} \).
3. \( f = 21.4 \) cm.

Teacher's Note:
1. The two lens positions are interchangeable because u and v swap their values.
2. Memorise \( f = \frac{D^2 - d^2}{4D} \) for such problems.

Answer: (a) \( \frac{f_1 f_2}{f_2 - f_1} \)

Explanation:
1. Powers add for lenses in contact: \( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{(-f_2)} = \frac{1}{f_1} - \frac{1}{f_2} \).
2. \( \frac{1}{f} = \frac{f_2 - f_1}{f_1 f_2} \), so \( f = \frac{f_1 f_2}{f_2 - f_1} \).
3. For a converging combination \( f_2 > f_1 \).

Teacher's Note:
1. Use signs carefully: the diverging lens has a negative focal length.
2. Options (b) and (c) give power, not focal length.

Answer: (c) become infinite

Explanation:
1. Lens maker's formula: \( \frac{1}{f} = \left(\frac{\mu_g}{\mu_l} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \).
2. If \( \mu_l = \mu_g \), the bracket \( \left(\frac{\mu_g}{\mu_l} - 1\right) = 0 \).
3. So \( \frac{1}{f} = 0 \) and \( f \to \infty \): the lens no longer bends light.

Teacher's Note:
1. Only the relative refractive index of the lens with respect to the surrounding matters.
2. An invisible glass rod in oil is the same idea.

Answer: (a) 0.3 s\( ^{-1} \)

Explanation:
1. Magnification \( m = \frac{f}{f + u} \). With the object distance x from the lens (u = -x), \( m = \frac{f}{f - x} = \frac{0.3}{0.3 - x} \).
2. \( \frac{dm}{dt} = \frac{0.3}{(0.3 - x)^2}\frac{dx}{dt} \). At x = 0.4 m, \( (0.3 - x)^2 = 0.01 \), so the factor is \( \frac{0.3}{0.01} = 30 \).
3. With \( \left|\frac{dx}{dt}\right| = 0.01 \) m/s, \( \left|\frac{dm}{dt}\right| = 30\times0.01 = 0.3 \) s\( ^{-1} \).

Teacher's Note:
1. Differentiate the magnification formula with respect to time using the chain rule.
2. The negative sign of m only tells us the image is inverted.

Answer: (d) 20/3 m

Explanation:
1. To the fish the flame appears higher: apparent height above the surface \( h' = \mu h = \frac{4}{3}\times2 = \frac{8}{3} \) m.
2. The fish is 4 m below the surface.
3. Apparent height from the fish \( = 4 + \frac{8}{3} = \frac{20}{3} \) m.

Teacher's Note:
1. Objects in the rarer medium look farther to an observer in the denser medium.
2. Add the fish depth to the apparent height above the surface.

Answer: (a) 80 cm

Explanation:
1. Place O at x = 0 with mirrors at x = +10 and x = -10 cm.
2. First images are at +20 and -20 cm. The second image in the mirror at x = -10 formed from the first image at +20 is at \( -20 - 30 = -40 \) cm. Similarly the other second image is at +40 cm.
3. Distance between them \( = 40 + 40 = 80 \) cm.

Teacher's Note:
1. Each new image is the reflection of the previous image in the opposite mirror.
2. Image distances grow in steps of 20 cm, the mirror separation.

Answer: (b) 20 times nearer

Explanation:
1. A telescope magnifies the visual angle by 20 times.
2. The tree is not enlarged in height, but its angular size becomes 20 times larger, as if the tree were 20 times closer.
3. So the tree appears 20 times nearer.

Teacher's Note:
1. Angular magnification of a telescope makes distant objects look nearer.
2. The actual height of the tree does not change.

Answer: (b) \( -3\hat{j} + 4\hat{k} \)

Explanation:
1. Along the normal (x-axis): \( v_{i,x} = 2v_{m,x} - v_{o,x} = 2(1) - 2 = 0 \).
2. Components parallel to the mirror stay equal to those of the object: \( v_{i,y} = -3 \) and \( v_{i,z} = 4 \).
3. So \( \vec{v}_i = -3\hat{j} + 4\hat{k} \).

Teacher's Note:
1. The image velocity along the normal depends on the mirror velocity, while the parallel components copy the object.
2. This follows from differentiating the image position.

Answer: (c) \( 15^\circ \)

Explanation:
1. At the first face the ray enters undeviated and hits the second face at an angle of incidence equal to the prism angle, \( 30^\circ \).
2. Snell's law at the second face: \( \sqrt{2}\sin30^\circ = \sin r \), so \( \sin r = \frac{1}{\sqrt{2}} \) and \( r = 45^\circ \).
3. Deviation \( = 45^\circ - 30^\circ = 15^\circ \).

Teacher's Note:
1. Normal incidence means no bending at the first surface.
2. Deviation equals the difference between the angles of emergence and incidence at the second face.

Answer: (a) red

Explanation:
1. Apparent rise \( = t\left(1 - \frac{1}{\mu}\right) \) increases with the refractive index.
2. Violet has the highest refractive index and red the lowest.
3. So red is raised the least.

Teacher's Note:
1. Shorter wavelengths have larger refractive index in glass.
2. Violet appears nearest to the top surface.

Answer: (a) \( \frac{x + y}{2} \)

Explanation:
1. For a real image \( |m| = \frac{f}{x - f} \), and for a virtual image \( |m| = \frac{f}{f - y} \).
2. Equal magnifications give \( x - f = f - y \).
3. So \( 2f = x + y \) and \( f = \frac{x + y}{2} \).

Teacher's Note:
1. The two cases correspond to the object beyond and inside the focal length.
2. The focal length is the mean of the two object distances.

Answer: (a) 14

Explanation:
1. For normal adjustment \( \frac{f_o}{f_e} = 10 \) and \( f_o + f_e = 1.1 \) m, so \( f_e = 0.1 \) m = 10 cm and \( f_o = 1 \) m.
2. With the image at the near point, \( M = \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right) \).
3. \( M = 10\left(1 + \frac{10}{25}\right) = 14 \).

Teacher's Note:
1. The magnification is a little larger when the final image is brought to 25 cm.
2. Normal adjustment means the final image is at infinity.

Answer: (a) at one of the foci, virtual and double its size

Explanation:
1. With \( u = -\frac{f}{2} \): \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{f} - \frac{2}{f} = -\frac{1}{f} \), so \( v = -f \).
2. The image is virtual, on the same side as the object, at distance f, which is at the focus on the object side.
3. Magnification \( m = \frac{v}{u} = \frac{-f}{-f/2} = 2 \).

Teacher's Note:
1. An object inside the focal length of a convex lens always gives a virtual, erect and magnified image.
2. This is how a simple magnifier works.

Answer: (d) \( \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) \)

Explanation:
1. Critical angle at the wall: \( \sin C = \frac{1}{\mu} = \frac{\sqrt{3}}{2} \), so \( C = 60^\circ \).
2. The wall is parallel to the axis, so the angle of refraction r at the end face satisfies \( r = 90^\circ - C = 30^\circ \).
3. Snell's law at the end: \( \sin\theta = \mu\sin r = \frac{2}{\sqrt{3}}\times\frac{1}{2} = \frac{1}{\sqrt{3}} \).

Teacher's Note:
1. The angle at the end face and the angle at the wall add up to 90 degrees.
2. Any larger angle of incidence would let the ray be totally reflected inside.

Answer: (a) 1 : 4

Explanation:
1. For a thin prism \( \delta = (\mu_{rel} - 1)A \).
2. In air \( \delta_a = 0.5A \). In water \( \mu_{rel} = \frac{1.5}{1.33} = 1.128 \), so \( \delta_w = 0.128A \).
3. \( \frac{\delta_w}{\delta_a} = \frac{0.128}{0.5} \approx 0.256 \approx \frac{1}{4} \).

Teacher's Note:
1. Use the refractive index of the prism relative to the surrounding medium.
2. A smaller index difference gives less bending.

Answer: (d) 11.0 cm

Explanation:
1. Objective (\( f_o = 1.5 \) cm, \( u = -2 \) cm): \( \frac{1}{v} = \frac{1}{1.5} - \frac{1}{2} = \frac{1}{6} \), so \( v_o = 6 \) cm.
2. Eyepiece (\( f_e = 6.25 \) cm, image at \( v = -25 \) cm): \( \frac{1}{u} = \frac{1}{v} - \frac{1}{f_e} = -\frac{1}{25} - \frac{1}{6.25} = -0.2 \), so the intermediate image is 5 cm from the eye lens.
3. Distance between the lenses \( = 6 + 5 = 11 \) cm.

Teacher's Note:
1. The image of the objective acts as the object for the eyepiece.
2. Use the correct signs for the virtual final image.

Answer: (c) Convex mirror

Explanation:
1. A convex mirror diverges the reflected rays, so it forms a small, erect, virtual image of a wide region.
2. This gives a larger field of view than a plane or concave mirror of the same size.
3. That is why convex mirrors are used as rear-view mirrors.

Teacher's Note:
1. Field of view is the region of space the mirror lets you see.
2. Convex mirrors trade image size for coverage.

Answer: (b) \( \sin^{-1}\left(\frac{2}{3}\right) \)

Explanation:
1. Frequency stays constant, so \( \frac{\mu_2}{\mu_1} = \frac{\lambda_1}{\lambda_2} = \frac{6000}{4000} = 1.5 \).
2. The second medium is denser. For a ray going from the second medium to the first, \( \sin C = \frac{\mu_1}{\mu_2} = \frac{2}{3} \).
3. So \( C = \sin^{-1}\left(\frac{2}{3}\right) \).

Teacher's Note:
1. Wavelength is inversely proportional to refractive index at a fixed frequency.
2. Total internal reflection needs the ray to start in the denser medium.

Answer: (b) 15

Explanation:
1. The convex lens focuses the parallel beam at its focus, 20 cm behind A.
2. For B to send out a parallel beam, this point must be the virtual focus of the concave lens, which lies 5 cm behind B.
3. So \( d = 20 - 5 = 15 \) cm.

Teacher's Note:
1. A pair of lenses that turns a parallel beam into a parallel beam is called afocal.
2. Here \( d = f_A - |f_B| \).

Answer: (c) 0.32 m

Explanation:
1. An upright image means the image is virtual, with \( m = -\frac{v}{u} = 5 \), so \( v = -5u \).
2. Mirror formula with \( f = -0.4 \) m: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \) gives \( \frac{1}{-5u} + \frac{1}{u} = \frac{4}{5u} = -\frac{1}{0.4} \).
3. \( u = -\frac{4\times0.4}{5} = -0.32 \) m, so the mirror is held 0.32 m from the face.

Teacher's Note:
1. A concave mirror gives an erect, magnified image only when the object is within the focal length.
2. The result 0.32 m is less than f = 0.4 m, as required.

Answer: (b) 5

Explanation:
1. Eyepiece magnification with the image at 25 cm: \( m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 6 \).
2. Total magnification \( M = m_o\times m_e \), so \( m_o = \frac{30}{6} \).
3. \( m_o = 5 \).

Teacher's Note:
1. The overall magnification is the product of the two lenses.
2. The eyepiece acts as a simple magnifier.

Answer: (d) 12 cm to the left of the lens

Explanation:
1. Light passes through the lens, reflects from the silvered plane face, and passes through the lens again.
2. Equivalent power \( = \frac{2}{f} \), so the system acts as a concave mirror of focal length \( \frac{15}{2} = 7.5 \) cm.
3. Mirror formula with \( u = -20 \) cm and \( f = -7.5 \) cm: \( \frac{1}{v} = -\frac{1}{7.5} + \frac{1}{20} = -\frac{5}{60} \), so \( v = -12 \) cm. The image is real and 12 cm to the left of the lens (on the object side).

Teacher's Note:
1. A silvered lens acts as a mirror with half the focal length.
2. The negative v shows the image is on the same side as the object.