CBSE Class 10 Physics Light Reflection and Refraction MCQs Set 18

Answer: (d) 60°

Explanation:
1. The glancing angle is measured from the surface, so the angle of incidence = 90° - 30° = 60°.
2. The angle of reflection is also 60°, and the deviation d = 180° - 2 × 60° = 60°.

Teacher's Note:
1. Glancing angle + angle of incidence = 90°.
2. Deviation on reflection = 180° - 2i.

Answer: (a) 4.4 × 10¹⁴ Hz

Explanation:
1. Speed in glass v = c/n = 3 × 10⁸/1.5 = 2 × 10⁸ m s^-1.
2. f = v/λ = 2 × 10⁸/(450 × 10^-9) = 4.4 × 10¹⁴ Hz.

Teacher's Note:
1. Frequency does not change from one medium to another.
2. Speed and wavelength both decrease in glass.

Answer: (d) 15 cm

Explanation:
1. Let the mirror be at 0, the object at -x and the image at +x.
2. After the move the mirror is at -5, the object-mirror distance is x - 10, so the image is at -5 + (x - 10) = x - 15.
3. The image shifts by x - (x - 15) = 15 cm.

Teacher's Note:
1. The image is as far behind the mirror as the object is in front.
2. Moving the mirror by d shifts the image by 2d.

Answer: (a) P → r; Q → p; R → q

Explanation:
1. A concave mirror can form real images (r).
2. A convex mirror always forms diminished virtual images (p).
3. A plane mirror forms virtual images with magnification one (q).

Teacher's Note:
1. Convex mirrors give a wide field of view.
2. Plane mirror: image the same size as the object.

Answer: (c) \( \frac{40}{3} \) m

Explanation:
1. For an object in air seen from water, apparent height = real height × μ_w.
2. = 10 × 4/3 = 40/3 m.

Teacher's Note:
1. Objects above the water appear taller to a diver.
2. Objects under water appear raised to an observer above.

Answer: (b) \( \frac{4}{\sqrt{3}} \) × 10⁸ m s^-1

Explanation:
1. sinC = n_B/n_A = v_A/v_B.
2. v_B = v_A/sin60° = 2 × 10⁸/(\( \sqrt{3} \)/2) = \( \frac{4}{\sqrt{3}} \) × 10⁸ m s^-1.

Teacher's Note:
1. Total internal reflection occurs from the denser (slower) to the rarer (faster) medium.

Answer: (a) convex lens

Explanation:
1. A convex lens forms real images for objects beyond its focus.
2. A concave lens, a convex mirror and a plane mirror (for real objects) form only virtual images.

Teacher's Note:
1. Real images can be received on a screen.

Answer: (b) 60°

Explanation:
1. n = sin[(A + D)/2]/sin(A/2), so \( \sqrt{3} \) = sin[(60° + D)/2]/sin30°.
2. sin[(60° + D)/2] = \( \sqrt{3} \)/2, so (60° + D)/2 = 60° and D = 60°.

Teacher's Note:
1. Use the prism formula at minimum deviation.

Answer: (d) black

Explanation:
1. A green object reflects only green light and absorbs the rest.
2. Red light is absorbed completely, so nothing is reflected and the dress looks black.

Teacher's Note:
1. Objects show only the colours they can reflect.

Answer: (b) -0.33

Explanation:
1. The far point is 3 m, so the lens must form the image of distant objects at 3 m: v = -3 m with u = ∞.
2. f = -3 m, so P = 1/f = -0.33 D (a concave lens).

Teacher's Note:
1. Myopia is corrected by a concave lens of focal length equal to the far point.

Answer: (a) a convex lens of power +1 D

Explanation:
1. For lenses in contact, powers add: P = +2 + (-1) = +1 D.
2. A positive power means a convex (converging) lens.

Teacher's Note:
1. Power of a combination = P₁ + P₂.

Answer: (b) C B A D

Explanation:
1. Compare the angles of deviation (C).
2. The greater deviation means a larger refractive index, so μ₁ > μ₂ (B).
3. Use n = c/v (A), which gives v₁ < v₂ (D).

Teacher's Note:
1. The more a colour is deviated, the slower it travels in glass.
2. Violet is deviated the most and red the least.

Answer: (c) 30°

Explanation:
1. d = 180° - 2i gives 120° = 180° - 2i.
2. i = 30°.

Teacher's Note:
1. Deviation on reflection = 180° - 2i.

Answer: (a) 450 nm

Explanation:
1. Speed in water = 3 × 10⁸/(4/3) = 2.25 × 10⁸ m s^-1.
2. λ = v/f = 2.25 × 10⁸/(5 × 10¹⁴) = 4.5 × 10^-7 m = 450 nm.

Teacher's Note:
1. The frequency stays the same; the wavelength shrinks by the factor n.

Answer: (a) 10 cm towards the object

Explanation:
1. If the mirror moves y towards the object, the new image distance is x - y behind the mirror.
2. The image position changes by 2y, so 2y = 20 cm and y = 10 cm towards the object.

Teacher's Note:
1. Moving a plane mirror by d moves the image by 2d.

Answer: (c) P → r; Q → q; R → r

Explanation:
1. A convex lens forms real images (r).
2. A concave lens always forms diminished virtual images (q).
3. A convex lens used as a magnifier (object within f) forms a magnified virtual image (r).

Teacher's Note:
1. The same lens can give real or virtual images depending on the object position.

Answer: (c) \( \frac{23}{2} \) m

Explanation:
1. Apparent depth of the bottom = 2/(4/3) = 1.5 m below the water surface.
2. Distance from the eyes = 10 + 1.5 = 11.5 = 23/2 m.

Teacher's Note:
1. Apparent depth = real depth / μ.
2. Add the height of the observer above the water.

Answer: (d) \( \sin^{-1}\frac{4}{5} \)

Explanation:
1. Q is the denser medium (lower speed), and total internal reflection occurs from Q to P.
2. sinC = n_P/n_Q = v_Q/v_P = 2/2.5 = 4/5.
3. C = sin^-1(4/5).

Teacher's Note:
1. sinC = (refractive index of rarer)/(refractive index of denser).

Answer: (d) plane mirror

Explanation:
1. A plane mirror forms only virtual images of real objects.
2. Reflection (concave mirror), refraction and a convex lens can all form real images.

Teacher's Note:
1. A real image can be taken on a screen.

Answer: (a) \( \sqrt{2} \)

Explanation:
1. n = sin[(A + D)/2]/sin(A/2) = sin₄₅°/sin₃₀°.
2. = (1/\( \sqrt{2} \))/(1/2) = \( \sqrt{2} \).

Teacher's Note:
1. Use the prism formula for minimum deviation.

Answer: (d) either blue or white

Explanation:
1. A blue object reflects blue light only.
2. It looks blue if blue light is present, which is the case in blue light and in white light.

Teacher's Note:
1. Red light is absorbed by a blue object, which would look black.

Answer: (a) myopia

Explanation:
1. A negative power means a concave lens.
2. A concave lens corrects myopia (short-sightedness).

Teacher's Note:
1. Hypermetropia needs a convex lens.