Page 246 - text book form physics kssm 2020

Solving Problems Involving Refraction of Light

Example 1

Figure 6.14 shows a light ray travelling from air into a
plastic block at an angle of 60°. The refractive index of 60°
plastic is 1.49. Air
Calculate: r Plastic
(a) angle of refraction, r.
(b) speed of light in plastic.

Solution: Figure 6.14

sin i c
(a) n = (b) n =
sin r v
sin i v = c
sin r =
n n
3.0 × 10 8
sin 60° =
= 1.49
1.49
8
= 2.01 × 10 m s –1
–1
r = sin sin 60°
1.49
= 35.54°

Example 2

When a coin inside a beaker containing a solution is observed, image of the coin is seen at a
2
height equal to of the depth of the solution. What is the refractive index of the solution?
7
Solution:
Solution
Based on Figure 6.15,
2
Apparent depth, h = H − H
7
5 2
–
= H h = H – H
7 7
H H
Refractive index of solution, n =
h Image of coin
H 2
7
= 5 – H
H
7
7 Coin
=
5 Figure 6.15
= 1.4

240
240 6.1.7
6.1.7