Page 395 - Schaum's Outline of Theory and Problems of Applied Physics

380 LIGHT [CHAP. 30

Now medium 1 is water and medium 2 is air. Hence
1.33
n 1 ◦
sinr = sin i = sin 40 = 0.855
n 2 1.00
r = 59 ◦
The angle of refraction is greater than the angle of incidence because n 2 < n 1 .

SOLVED PROBLEM 30.15

◦
A beam of light strikes a pane of glass at an angle of incidence of 50 . If the angle of refraction is 30 ,
◦
find the index of refraction of the glass.
According to Snell’s law, n 1 sin i = n 2 sin r. Here air is medium 1, so n 1 = 1.00 and

sin i sin 50 ◦
n 2 = (n 1 ) = (1.00) = 1.53
sinr sin 30 ◦
SOLVED PROBLEM 30.16
◦
A ray of light enters a glass plate whose index of refraction is 1.5 at an angle of incidence of 50 (Fig. 30-9).
At what angle does the ray leave the other side of the plate?

i 1

Air
Glass

r 1
i 2

Glass
Air

r 2

Fig. 30-9

We start by finding the angle of refraction r 1 at the first side of the plate:

n air 1.0
◦
sinr 1 = (sin i 1 ) = (sin 50 )
n glass 1.5
1.0
r 1 = sin −1 sin 50 ◦ = sin −1 0.51 = 30.7 ◦
1.5
The angle of incidence i 2 at the other side of the plate is equal to r 1 . Hence

n glass n glass 1.5 ◦
sinr 2 = (sin i 2 ) = (sinr 1 ) = (sin 30.7 )
n air n air 1.0
1.5
r 2 = sin −1 sin 30.7 ◦ = 50 ◦
1.0
The ray leaves the glass plate parallel to its original direction but shifted to one side. This result would be true
regardless of the index of refraction of the glass.