CBSE_NCERT Class X Ch.10 – Light - Reflection and Refraction Solved Think Zone

Class X NCERT Solution, MCQ's Study Notes, Q & A Chapter 10 – Light -Reflection and Refraction

Q.1.Why does a light ray incident on a rectangular glass slab immersed in any medium emerge parallel to itself?

Ans: The extent of bending of the ray of light at the opposite parallel faces AB (air-glass interface) and CD (glass-air interface) of the rectangular glass slab is equal and opposite. This is why the ray emerges parallel to the incident ray. OR,

(The glass slab is rectangular and both sides of the glass slab have the same medium. The light refracts in such a way that incident and emergent rays are parallel.)

Q.2.An object is placed at a distance of 10cm from a convex mirror of focal length of 15cm . Find the position and nature of the image

Solution: Given : f = +15cm u = -10cm v = ?’

1/ f = 1/u +1/v

1/15 = -1/10 + 1/v

1/15 + 1/10 = 1/v

1/v = 1/6 v = 6cm

Position of the image: Since V is formed at a distance of 6cm away on right side of the mirror .

Nature of the image : Since the mirror is convex , image is virtual , erect and small sized.

Q.3.A concave lens of focal length 15 cm , forms an image 10 cm from the lens. How far is the object

Solution : Given: F = -15cm V = -10cm U = ?

1/f = 1/v - 1/u

1/-15 = 1/-10 - 1/u

1/u = - 1/10 + 1/15

1/u = -1/30 U = -30cm

The object should be placed at a distance of 30cm in front of the mirror

Q.4. An object 5.0cm in length is pl aced at a distance of 20cm in front of a convex mirror of radius of Curvature 30cm . F ind the position of the image , its nature and size.

Solution: Given:

Size of the object [O] = +15 cm

R = +30 cm

F = R/2 = +15cm

U = -20cm

V = ?

1/f = 1/v + 1/u

1/15 = 1/v-1/20

1/v = 1/15 +1/20

1/v = 7/60

v = 60/7= 8.6cm

Position of the image : Since v is +ve and image formed at a distance of 8.6 cm back side of the mirror [right side of the mirror]

Size of the image: m = hI/hO = -v/u

= I/5 = -60 / 7x 1/20

I = 15/7= 2.2cm

I [ size of the image] 2.2cm

Nature of the image: Since the mirror is convex image is virtual , erect and small sized.

Q.5. The density of the atmosphere decreases with height, as does the index of refraction. Explain how one can see the sun after it has set. Why does the setting sun appear flattened?

Answer: The change in atmospheric density results in refraction of the light from the sun, bending it toward the earth.
Consequently, the sun can be seen even after it is just below the horizon. Also, the light from lower portion of the sun is refracted more than that from the upper portion, so the lower part appears to be slightly higher in the sky. The effect is an apparent flattening of the disk into an ellipse.

Q. 6. How does a thin layer of water on the road affect the light you see reflected off the road from your own headlights? How does it affect the light you see reflected from the headlights of an oncoming car?

Answer: The layer of water greatly reduces the light reflected back from the car’s headlights, but increases the light reflected by the road of light from the headlights of oncoming cars.

Q.5.Choose the correct alternative in Question 1-5

1. The image formed by a concave mirror is observed to be virtual, erect and larger than the object Then the position of the object should be

[A] Between the focus and center of curvature.

[b] At the centre of curvature.

[c] Beyond the centre of curvature.

[d] Between the pole of the mirror and its focus. Ans: [ d]

2. Where should an object be placed, so that a real and inverted image of the same size is obtained by a Convex lens?

[a] at the focus of the lens.

[b] at twice the focal length.

[c] at infinity.

[d] between the optical centre of the lens and its focus. Ans: [ b ]

3. A spherical mirror and a thin spherical lens have each a focal length of -15cm.