CBSE Class 10 Physics Light Reflection and Refraction MCQs Set 20

Answer: (d) the aperture setting of the camera

Explanation:
1. The amount of light is controlled by the opening (aperture) of the lens.
2. A larger aperture lets in more light.

Teacher's Note:
1. The f-number (aperture setting) controls exposure.

Answer: (a) remain unchanged

Explanation:
1. The focal length of a lens depends on its curvature and refractive index, not on its aperture.
2. Covering part of the lens leaves the focal length unchanged.

Teacher's Note:
1. Only brightness is reduced.

Answer: (c) the intensity of the image decreases

Explanation:
1. Rays from every point of the object pass through the lower half of the lens too.
2. The image position stays the same; only the intensity decreases.

Teacher's Note:
1. Cutting a lens does not move the image of an on-axis point.

Answer: (d) must not be less than 10 cm

Explanation:
1. For a real image, \( u + v \ge 4f \) in magnitude.
2. If no position of the lens gives an image, \( 4f > 40 \), so \( f > 10 \) cm.

Teacher's Note:
1. The minimum object-to-image distance is 4f.

Answer: (d) the image is always virtual and diminished

Explanation:
1. A convex lens forms real images as well as virtual images and may magnify or diminish.
2. The image is not always virtual and diminished.

Teacher's Note:
1. Image nature depends on the object distance.

Answer: (d) 1.428

Explanation:
1. \( v = 0.7c \), so \( n = c/v = 1/0.7 \).
2. \( n \approx 1.428 \).

Teacher's Note:
1. Refractive index = speed in vacuum ÷ speed in the medium.

Answer: (b) the centre of curvature

Explanation:
1. Magnification −1 means an image equal in size to the object and inverted (real).
2. For a concave mirror this happens when the object is at the centre of curvature.

Teacher's Note:
1. Image also forms at C.

Answer: (b) 11

Explanation:
1. \( n = \dfrac{360^\circ}{30^\circ} - 1 = 12 - 1 = 11 \).

Teacher's Note:
1. Use \( n = 360/\theta - 1 \).

Answer: (b) 1.67

Explanation:
1. \( v = 0.6c \), so \( n = 1/0.6 \approx 1.67 \).

Teacher's Note:
1. A 40% decrease leaves 60% of the original speed.

Answer: (c) a plane mirror

Explanation:
1. A plane mirror always gives an erect, virtual image of the same size.
2. So m = +1 only for a plane mirror.

Teacher's Note:
1. Concave mirrors give +1 only in the limit of an object at the pole.

Answer: (b) critical angle

Explanation:
1. Rays going from a denser to a rarer medium bend away from the normal.
2. At the critical angle of incidence, the refracted ray travels along the interface (angle of refraction 90°).

Teacher's Note:
1. Beyond the critical angle there is total internal reflection.

Answer: (b) greater than 1

Explanation:
1. Refractive index = c ÷ v, and v is less than c in all material media.
2. So the absolute refractive index of any medium is greater than 1.

Teacher's Note:
1. Vacuum has n = 1 exactly.

Answer: (c) a convex lens of focal length 0.20 m

Explanation:
1. \( f = 1/P = 1/5 = 0.20 \) m.
2. Positive power means a convex (converging) lens.

Teacher's Note:
1. f in metres is 1/P.

Answer: (c) violet light

Explanation:
1. Violet light is slowed the most in glass.
2. So the refractive index is highest for violet.

Teacher's Note:
1. Cauchy: n decreases as wavelength increases.

Answer: (c) concave mirror

Explanation:
1. A concave mirror gives a real, enlarged image when the object is between F and C.
2. A convex mirror and a plane mirror never give real images.

Teacher's Note:
1. Only concave mirrors (and convex lenses) form real images.

Answer: (d) more or less than 1 depending on object position

Explanation:
1. Convex lens: m depends on object position.
2. It can be greater than, equal to or less than 1.

Teacher's Note:
1. m = v/u.

Answer: (a) retraces its path

Explanation:
1. A ray along the radius hits the mirror normally.
2. It is reflected back along the same path.

Teacher's Note:
1. The normal at any point passes through the centre of curvature.

Answer: (b) 10 cm

Explanation:
1. For a concave lens \( f=-15 \) cm and a diminished virtual image with \( v = u/3 \) (same side).
2. \( \dfrac1v-\dfrac1u = -\dfrac1{15} \) gives \( -\dfrac2u = -\dfrac1{15} \) with u numerically 30 cm, so \( |v| = 10 \) cm.

Teacher's Note:
1. A concave lens always forms a virtual, diminished image.

Answer: (c) 45°

Explanation:
1. \( \sin C = 1/\mu = 1/\sqrt2 \).
2. \( C = 45^\circ \).

Teacher's Note:
1. Critical angle: sin C = 1/μ.

Answer: (a) glass

Explanation:
1. Sound is a mechanical wave and travels fastest in solids.
2. It cannot travel in vacuum.

Teacher's Note:
1. Speed: solid > liquid > gas.

Answer: (b) presbyopia

Explanation:
1. Presbyopia is the loss of the eye's power of accommodation with age.

Teacher's Note:
1. Corrected by bifocal lenses.

Answer: (c) optical centre

Explanation:
1. The optical centre is the point through which a ray goes without any deviation.

Teacher's Note:
1. Rays through the focus emerge parallel.

Answer: (b) refraction

Explanation:
1. Light from the bottom of the pool bends away from the normal on leaving water.
2. The pool appears shallower because of this refraction.

Teacher's Note:
1. Apparent depth = real depth ÷ refractive index.

Answer: (a) 4 m

Explanation:
1. The image is as far behind the mirror as the object is in front, i.e. 4 m.

Teacher's Note:
1. The distance between person and image is 8 m.

Answer: (c) pq = f²

Explanation:
1. Newton's formula: \( (u-f)(v-f) = f^2 \) for a mirror, so \( pq = f^2 \).

Teacher's Note:
1. p and q are measured from the focus.

Answer: (a) (u₁ + u₂)/2

Explanation:
1. Real image: \( m = \dfrac{f}{u_1-f} \). Virtual image: \( m = \dfrac{f}{f-u_2} \) (magnitudes).
2. Equal magnitudes give \( u_1 - f = f - u_2 \), so \( f = (u_1+u_2)/2 \).

Teacher's Note:
1. Take magnitudes of u from the lens.

Answer: (b) frequency does not change but wavelength changes

Explanation:
1. Frequency is fixed by the source and does not change at a boundary.
2. The speed and the wavelength both decrease in glass.

Teacher's Note:
1. \( \lambda = v/\nu \).

Answer: (a) 1.125

Explanation:
1. \( n_{gw} = n_g/n_w = (3/2)/(4/3) = 9/8 = 1.125 \).

Teacher's Note:
1. Relative index = ratio of absolute indices.

Answer: (d) 2

Explanation:
1. \( \dfrac1v = \dfrac1f - \dfrac1u \) gives \( \dfrac1v = \dfrac1{30}-\dfrac1{15}=-\dfrac1{30} \), so \( v = -30 \) cm (virtual).
2. \( m = v/u = -30/-15 = 2 \).

Teacher's Note:
1. Object inside the focal length gives a magnified virtual image.

Answer: (a) sin⁻¹(4/5)

Explanation:
1. \( \sin C = n_w/n_g = (4/3)/(5/3) = 4/5 \).
2. \( C = \sin^{-1}(4/5) \).

Teacher's Note:
1. Critical angle: sin C = n(rarer)/n(denser).