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question no. 4,5,9
Answers
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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