Question

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question no. 4,5,9​

Answer 1

4) n(glass) = 3/2

n(water) = 4/3

n(glass w.r.t water) = n(glass) / n(water)

= 3/2 / 4/3

= 3/2 × 3/4

= 9/8

(b) is the correct answer.

5) height of the object, h = 1.2 m = 120cm

focal length, f = – 20 cm

Image, v = – 60cm

using mirror formula

1/f = 1/v + 1/u

1/ –20 = 1/– 60 + 1 / u

–1/20 + 1/60 = 1/u

( –3+1 ) / 60 = 1 / u

– 2 / 60 = 1/ u

– 1 / 30 = 1 / u

Reciprocal both sides

– 30 cm = u

Also,

– V / u = h'/ h

( where h'= height of the image

h = height of the object )

60 / – 30 = h'/ 120

– 2 = h'/ 120

h'= 240 cm

Ans: Height of the image formed is 2.4 m and the image distance from mirror is 30cm to the left.

9) Given magnification, m = – 1

distance of object, u = –30 cm

we know, for mirrors

m = –v/u

– 1 = –v / – 30

– 1 = v / 30

– 30cm = v

Image distance is 30cm to the left of mirror.

We know that when an object is placed at centre of curvature it's image is formed again at R and since the image is real and inverted.

Hence, it is a concave mirror.

R=2f

– 30 = 2f

– 15 cm = f

Focal length of the mirror is 15cm

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Answer 2

Explanation:

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