CBSE Class 10 Physics Light Reflection and Refraction MCQs Set 19

Answer: (c) a large number of reflected images

Explanation:
1. Two parallel plane mirrors reflect the image of each other repeatedly.
2. So an infinite (very large) number of images is formed.

Teacher's Note:
1. Images get fainter with each reflection.

Answer: (b) 100°

Explanation:
1. Angle of reflection = angle of incidence = 50°.
2. Angle between incident and reflected rays = 50° + 50° = 100°.

Teacher's Note:
1. The angle between the two rays is twice the angle of incidence.

Answer: (c) half of one's height

Explanation:
1. The image is as far behind the mirror as the person is in front.
2. Rays from the head and feet need a mirror of only half the person's height.

Teacher's Note:
1. The result does not depend on the distance from the mirror.

Answer: (b) 30°

Explanation:
1. Angle with the mirror 15° gives angle of incidence 90° − 15° = 75°.
2. Deviation = 180° − 2i = 180° − 150° = 30°.

Teacher's Note:
1. Deviation by a plane mirror is 180° − 2i.

Answer: (c) S + d

Explanation:
1. The image is at distance d behind the mirror.
2. The person is S in front of the mirror, so the distance to the image is S + d.

Teacher's Note:
1. Measure both distances from the mirror along the same line.

Answer: (b) 5

Explanation:
1. \( n = \dfrac{360^\circ}{\theta} - 1 \) (when 360/θ is an integer).
2. \( n = 360/60 - 1 = 5 \).

Teacher's Note:
1. A smaller angle between the mirrors gives more images.

Answer: (b) 36°

Explanation:
1. \( n = \dfrac{360^\circ}{\theta} - 1 = 9 \) gives \( 360/\theta = 10 \).
2. \( \theta = 36^\circ \).

Teacher's Note:
1. Use \( n = 360/\theta - 1 \) for an even-integer ratio.

Answer: (c) f

Explanation:
1. Reflection changes only the direction of the ray; the focusing property comes from the surface shape, not from the medium.
2. So the focal length of a mirror does not change when it is immersed in a liquid.

Teacher's Note:
1. Lenses change focal length in a liquid, mirrors do not.

Answer: (a) a convergent lens

Explanation:
1. \( \dfrac1f = (\mu_{rel}-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right) \), where \( \mu_{rel} = 1.5/1.33 > 1 \).
2. Glass is optically denser than water, so the lens still converges the light (with a longer focal length).

Teacher's Note:
1. A lens becomes diverging only if the surrounding medium is denser than the lens.

Answer: (b) 20 cm

Explanation:
1. For a plano-convex lens, \( \dfrac1f = (\mu-1)\dfrac1R = 0.5\times\dfrac1{10} \).
2. \( f = 20 \) cm.

Teacher's Note:
1. Plano-convex: only one curved surface of radius R.

Answer: (d) 9 D

Explanation:
1. \( P_1 = 1/0.20 = 5 \) D and \( P_2 = 1/0.25 = 4 \) D.
2. Power of the combination = 5 + 4 = 9 D.

Teacher's Note:
1. Powers add for lenses in contact.

Answer: (c) the full image is formed but it is less bright

Explanation:
1. Every part of the lens forms a complete image of the object.
2. Covering the centre removes some of the rays, so the image is complete but dimmer.

Teacher's Note:
1. Blocking part of a lens reduces brightness, not the image shape.

Answer: (a) −60 cm, −5/3 D

Explanation:
1. \( P = \dfrac1{0.20} - \dfrac1{0.15} = 5 - 6.67 = -\dfrac53 \) D.
2. \( f = 1/P = -0.6 \) m = −60 cm.

Teacher's Note:
1. A diverging combination has negative focal length and power.

Answer: (a) 1 and 3 only

Explanation:
1. Total internal reflection requires light going from a denser to a rarer medium.
2. The angle of incidence must exceed the critical angle.

Teacher's Note:
1. Both conditions must hold together.

Answer: (c) stars

Explanation:
1. A microscope magnifies tiny objects that are close, such as bacteria, cells and viruses.
2. Stars are very distant and are viewed with a telescope.

Teacher's Note:
1. Microscope: small and near; telescope: large and far.

Answer: (b) persistence of vision

Explanation:
1. The eye retains an image for about 1/16 s after the light is gone.
2. Frames shown at 16 or more per second blend into smooth motion.

Teacher's Note:
1. This is persistence of vision.

Answer: (c) a horse with less brightness

Explanation:
1. Light from every part of the horse passes through the unpainted parts of the lens.
2. So the image is complete but gets less light, hence is dimmer.

Teacher's Note:
1. Covering part of a lens only lowers the brightness.

Answer: (a) I1 = I2

Explanation:
1. Intensity at a minimum is proportional to \( (\sqrt{I_1}-\sqrt{I_2})^2 \) and at a maximum to \( (\sqrt{I_1}+\sqrt{I_2})^2 \).
2. Contrast is best when the minimum is zero, i.e. \( I_1 = I_2 \).

Teacher's Note:
1. Equal intensities give perfectly dark fringes.

Answer: (c) increases

Explanation:
1. \( \mu_{rel} = \mu_{glass}/\mu_{water} \) is smaller than \( \mu_{glass}/\mu_{air} \).
2. The focal length depends inversely on \( \mu_{rel}-1 \), so it increases.

Teacher's Note:
1. A lens is weaker in water than in air.

Answer: (a) 0°

Explanation:
1. For normal incidence the angle of incidence is 0°.
2. The ray goes straight, so the angle of refraction is 0°.

Teacher's Note:
1. No bending for normal incidence.

Answer: (d) 11

Explanation:
1. \( m = 1 + \dfrac{D}{f} = 1 + \dfrac{25}{2.5} \).
2. \( m = 11 \).

Teacher's Note:
1. Image at the least distance of distinct vision, D = 25 cm.

Answer: (b) decreased

Explanation:
1. Every part of the lens forms the whole image.
2. Removing half the lens halves the light, so the image becomes dimmer.

Teacher's Note:
1. Intensity decreases; position and size are unchanged.

Answer: (a) 4000 Å

Explanation:
1. In vacuum \( \lambda_0 = c/\nu = 3\times10^8/5\times10^{14} = 6\times10^{-7} \) m = 6000 Å.
2. In the medium \( \lambda = \lambda_0/\mu = 6000/1.5 = 4000 \) Å.

Teacher's Note:
1. Frequency is constant; wavelength is divided by μ.

Answer: (c) biconvex

Explanation:
1. A camera needs to form a real image on the film or sensor.
2. A biconvex (convex) lens forms real images.

Teacher's Note:
1. Camera: convex lens; the eye also works like a camera.

Answer: (c) 6 × 10¹⁴ Hz

Explanation:
1. \( \nu = c/\lambda = 3\times10^8/5\times10^{-7} \).
2. \( \nu = 6\times10^{14} \) Hz.

Teacher's Note:
1. Convert Å to metres: 1 Å = 10⁻¹⁰ m.

Answer: (b) concave

Explanation:
1. A concave mirror held close to a tooth forms an enlarged, erect, virtual image.
2. Hence the dentist uses a concave mirror.

Teacher's Note:
1. Concave mirrors magnify when the object is within the focal length.

Answer: (b) 1.6

Explanation:
1. \( v_1 = \dfrac{3.25+0.75}{2}\times10^8 = 2.0\times10^8 \) m/s and \( v_2 = 1.25\times10^8 \) m/s.
2. Refractive index of medium 2 with respect to medium 1 = \( v_1/v_2 = 2.0/1.25 = 1.6 \).

Teacher's Note:
1. Relative refractive index = speed in the first medium ÷ speed in the second.

Answer: (b) focal length

Explanation:
1. A lens converges or diverges light according to its curvature, which sets the focal length (power = 1/f).
2. The focal length is therefore the quantity that decides the ability.

Teacher's Note:
1. Power is the reciprocal of the focal length in metres.

Answer: (d) between F and the optical centre

Explanation:
1. A convex lens forms a virtual image only when the object is between the optical centre and the principal focus.
2. For an object at F the image is at infinity, and beyond F it is real.

Teacher's Note:
1. Virtual, erect, magnified image: object inside the focal length.