CBSE Class 12 Physics Ray Optics and Optical Instruments MCQs Set 06

Answer: (c) 60 cm in front of the mirror, real, inverted and 3 times the size of the object

Explanation:
1. Cartesian signs: u = −20 cm, f = −15 cm.
2. \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{15} + \frac{1}{20} = -\frac{1}{60} \), so v = −60 cm (in front of the mirror, real).
3. m = \( -\frac{v}{u} = -\frac{-60}{-20} \) = −3, so the image is inverted and 3 times the size.
4. Answer: 60 cm in front of the mirror, real, inverted and 3 times the size

Teacher's Note:
1. The object lies between F and C, so the image must lie beyond C.
2. A negative m always means a real, inverted image for a single mirror.

Answer: (a) a virtual, erect and magnified image

Explanation:
1. For an object between P and F of a concave mirror, the reflected rays diverge and appear to come from a point behind the mirror.
2. So the image is virtual, erect and magnified, which lets the dentist see a large, upright view of the tooth.

Teacher's Note:
1. The same property makes a concave mirror useful as a shaving mirror.
2. A convex mirror always diminishes, so it cannot be used for this purpose.

Answer: (b) Its frequency

Explanation:
1. The frequency of light is fixed by the source, so it stays the same in every medium.
2. Speed falls to c/n and wavelength falls to λ/n; an oblique ray also bends.
3. The quantity that does not change is the frequency.

Teacher's Note:
1. Colour depends on frequency, which is why light does not change colour on entering glass.
2. Speed and wavelength both decrease by the same factor n.

Answer: (d) 5 cm

Explanation:
1. Apparent depth = \( \frac{\text{Real depth}}{n} = \frac{20}{4/3} \) = 15 cm
2. Apparent rise = 20 − 15 = 5 cm

Teacher's Note:
1. 15 cm is the apparent depth, not the apparent rise; read the question carefully.
2. The shift can also be found directly as d(1 − 1/n).

Answer: (b) A convex lens used as a magnifying glass

Explanation:
1. Optical fibres and endoscopes guide light by repeated total internal reflection; a periscope prism reflects light totally at its hypotenuse face.
2. A magnifying glass works by refraction through a convex lens. So the answer is a convex lens used as a magnifying glass.

Teacher's Note:
1. An endoscope is a bundle of optical fibres.
2. Total internal reflection needs light going from a denser to a rarer medium at more than the critical angle.

Answer: (b) 1.41

Explanation:
1. The ray enters AB normally, so it is not bent and meets AC at an angle of incidence of 45°.
2. For total internal reflection, 45° must be at least the critical angle, so sin 45° ≥ \( \frac{1}{n} \).
3. n ≥ \( \frac{1}{\sin 45^\circ} = \sqrt{2} \) ≈ 1.41

Teacher's Note:
1. Ordinary crown glass (n = 1.5) satisfies this, which is why such prisms work in periscopes.
2. After reflection the ray leaves face BC normally, turned through 90°.

Answer: (c) 90 cm inside the glass

Explanation:
1. \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \), with n₁ = 1, n₂ = 1.5, u = −30 cm, R = +10 cm.
2. \( \frac{1.5}{v} + \frac{1}{30} = \frac{0.5}{10} \) ⇒ \( \frac{1.5}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} \)
3. v = +90 cm, so the image is real and 90 cm inside the glass.

Teacher's Note:
1. R is positive because the centre of curvature lies inside the glass, on the side of the outgoing light.
2. Remember to multiply n₂ by v, not divide it.

Answer: (d) 80 cm

Explanation:
1. In air: \( \frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} + \frac{1}{20}\right) \) = \( \frac{1}{20} \), so f = 20 cm.
2. In water: \( \frac{1}{f_w} = \left(\frac{1.5}{4/3} - 1\right)\left(\frac{2}{20}\right) = 0.125 \times 0.1 \) = \( \frac{1}{80} \)
3. f_w = 80 cm (still converging)

Teacher's Note:
1. The lens is still converging because glass is optically denser than water.
2. Use the refractive index of the lens relative to the surrounding medium.

Answer: (a) virtual, erect, 10 cm from the lens on the object side, one-third the size

Explanation:
1. u = −30 cm, f = −15 cm. \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = -\frac{1}{15} - \frac{1}{30} = -\frac{1}{10} \), so v = −10 cm.
2. m = \( \frac{v}{u} = \frac{-10}{-30} = \frac{1}{3} \)
3. The image is virtual, erect, 10 cm from the lens on the object side and one-third the size.

Teacher's Note:
1. A concave lens always forms a virtual, erect, diminished image of a real object.
2. The lens formula has a minus sign (1/v − 1/u), unlike the mirror formula.

Answer: (c) convex lens of focal length 66.7 cm

Explanation:
1. P = P₁ + P₂ = 4 − 2.5 = +1.5 D
2. F = \( \frac{100}{1.5} \) cm ≈ 66.7 cm; positive power means converging.
3. Answer: convex lens of focal length 66.7 cm

Teacher's Note:
1. Powers add directly; focal lengths do not.
2. Use F (cm) = 100/P (D).

Answer: (b) 1.41

Explanation:
1. n = \( \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin 45^\circ}{\sin 30^\circ} \)
2. n = \( \frac{0.707}{0.5} \) = √2 ≈ 1.41

Teacher's Note:
1. Halve (A + δm) before taking the sine.
2. At minimum deviation, i = (A + δm)/2 = 45° here.

Answer: (d) is parallel to the base

Explanation:
1. At minimum deviation r₁ = r₂ = A/2, so the ray passes symmetrically through the prism.
2. In an isosceles (or equilateral) prism this symmetric path is parallel to the base.

Teacher's Note:
1. Also at minimum deviation, i = e.
2. This setting is used in the laboratory to find the refractive index of a prism.

Answer: (a) 75

Explanation:
1. m = \( \frac{L}{f_o} \times \frac{D}{f_e} \) = \( \frac{15}{1} \times \frac{25}{5} \)
2. m = 15 × 5 = 75

Teacher's Note:
1. Both lenses of a microscope have short focal lengths, which makes m large.
2. With the final image at D, use (1 + D/fₑ) instead of D/fₑ.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Rear-view mirrors are convex, so the Assertion is correct.
2. A convex mirror forms an erect, diminished, virtual image of objects spread over a large area, so the Reason is correct.
3. The erect image and wide field of view are exactly why it is chosen, so the Reason explains the Assertion.

Teacher's Note:
1. Objects in a convex mirror look farther away than they are, because the image is small.
2. A plane mirror would show a much smaller area behind the vehicle.

Answer: (c) A is true, but R is false.

Explanation:
1. Optical fibres do carry signals with very little loss, so the Assertion is correct.
2. Light is guided by repeated total internal reflection at the core–cladding boundary, not by refraction. The Reason is false.

Teacher's Note:
1. The core has a higher refractive index than the cladding.
2. Almost no light escapes at each reflection, which keeps the loss small.

Answer: (d) A is false, but R is true.

Explanation:
1. P = \( \frac{1}{f} = \frac{1}{0.5\,\text{m}} \) = +2 D, not +0.5 D. The Assertion is false.
2. The Reason correctly defines power, P = 1/f (f in m), so it is true.

Teacher's Note:
1. Using f = 50 (in cm) directly in 1/f is a common mistake.
2. Use P = 100/f when f is in cm.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Frequency is decided by the source and does not change on refraction. The Assertion is true.
2. v = c/n, and n > 1 for glass, so light is slower in glass. The Reason is true.
3. The fall in speed explains the change in wavelength, not why frequency stays the same, so the Reason is not the correct explanation.

Teacher's Note:
1. v = fλ: when v falls with f fixed, λ must fall.
2. Choose (b) when both are true but the Reason answers a different question.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( \frac{1.5}{1.63} - 1 \) is negative, so f becomes negative and the lens diverges. The Assertion is true.
2. The Reason states exactly this sign change, and it is true.
3. The negative factor is the reason the convex lens becomes diverging, so the Reason explains the Assertion.

Teacher's Note:
1. The lens shape is unchanged; only its refractive index relative to the medium matters.
2. An air bubble in water behaves as a diverging lens for the same reason.

Answer: (d) A is false, but R is true.

Explanation:
1. m = f_o/f_e, so decreasing f_o decreases m. The Assertion is false.
2. The Reason gives the correct formula, so it is true.

Teacher's Note:
1. A telescope needs a long-focus objective and a short-focus eyepiece.
2. This is the opposite of a microscope, whose objective has a very short focal length.

Answer: (b) +5 D

Explanation:
1. Light from a distant window is focused at the focal point, so f = 20 cm = 0.2 m.
2. P = \( \frac{1}{0.2} \) = +5 D

Teacher's Note:
1. Focusing a distant object is a quick way to find the focal length of a convex lens.
2. Power is positive for a converging lens.

Answer: (a) 60 cm from the lens; the image is real, inverted and twice the size of the flame

Explanation:
1. u = −30 cm, f = +20 cm. \( \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} \), so v = +60 cm.
2. m = \( \frac{v}{u} = \frac{60}{-30} \) = −2
3. Answer: 60 cm from the lens; real, inverted and twice the size

Teacher's Note:
1. An image that can be caught on a screen is always real.
2. The candle lies between F and 2F, so the image is beyond 2F and magnified.

Answer: (d) be complete but less bright

Explanation:
1. Every part of the lens receives light from every point of the flame and brings it to the same image point.
2. So the uncovered half still forms the complete image, but with less light: complete but less bright.

Teacher's Note:
1. Only the amount of light reaching the image is halved.
2. The image position depends on f and u, which are unchanged.

Answer: (c) 20 cm

Explanation:
1. For an equiconvex lens, R₁ = R and R₂ = −R, so \( \frac{1}{f} = (n - 1)\frac{2}{R} \).
2. R = 2(n − 1)f = 2 × 0.5 × 20 = 20 cm

Teacher's Note:
1. For n = 1.5, an equiconvex lens has R = f.
2. Watch the sign of R₂ for the second surface.

Answer: (b) 60 cm, converging

Explanation:
1. \( \frac{1}{F} = \frac{1}{20} + \frac{1}{-30} = \frac{3 - 2}{60} = \frac{1}{60} \)
2. F = +60 cm, so the combination is converging with focal length 60 cm.

Teacher's Note:
1. In power terms: +5 D − 3.33 D = +1.67 D.
2. The stronger lens decides whether the combination converges or diverges.

Answer: (c) 30

Explanation:
1. m = \( \frac{f_o}{f_e} = \frac{120}{4} \) = 30

Teacher's Note:
1. Take the ratio of the objective to the eyepiece focal length, not the other way round.
2. Both focal lengths must be in the same unit.

Answer: (a) 124 cm

Explanation:
1. In normal adjustment, the focus of the eyepiece coincides with the focus of the objective.
2. L = f_o + f_e = 120 + 4 = 124 cm

Teacher's Note:
1. 116 cm (the difference) is a common slip.
2. The tube must be a little longer than the objective's focal length.

Answer: (d) 34.8

Explanation:
1. m = \( \frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right) = 30\left(1 + \frac{4}{25}\right) \)
2. m = 30 × 1.16 = 34.8

Teacher's Note:
1. Forming the image at D gives a slightly larger magnification, but the eye is strained.
2. Normal adjustment is preferred for long viewing.

Answer: (b) it collects more light, so faint objects appear brighter

Explanation:
1. A larger aperture admits more light from a distant object into the telescope.
2. So the image of faint stars becomes brighter: it collects more light.

Teacher's Note:
1. Aperture and focal length are independent choices.
2. The final image of an astronomical telescope stays inverted.

Answer: (a) a small convex secondary mirror, through a hole in the primary mirror

Explanation:
1. In the Cassegrain design, a small convex mirror reflects the converging light back towards the primary mirror.
2. The light passes through a central hole in the primary to the eyepiece: a small convex secondary mirror, through a hole in the primary.

Teacher's Note:
1. Mirror objectives have no chromatic aberration and can be supported over their whole back.
2. The secondary mirror also makes the instrument shorter.