Page 44 - Physics 10_Float

GEOMETRICAL OPTICS

Example 12.3: A ray of light enters from air into glass. The
o
angle of incidence is 30 . If the refractive index of glass is 1.52,
then find the angle of refraction ‘r’.
o
Solution: Given that, i = 30 , n= 1.52
Using Snell's law, sin i = n
sin r
1.52 sin r = sin 30 o
o
or sin r = sin 30 /1.52
sin r = 0.33
-1
r = sin (0.33)
r = 19.3 o Normal Refracted ray
o
Hence, angle of refraction is 19.3 . N
12.5 TOTAL INTERNAL REFLECTION r
When a ray of light travelling in denser medium enters into a Air
Glass
rarer medium, it bends away from the normal (Fig.12.9-a). If
i
the angle of incidence ‘i’ increases, the angle of refraction ‘r’ Incident
also increases. For a particular value of the angle of ray
o
incidence, the angle of refraction becomes 90 . The angle of i > c
incidence, that causes the refracted ray in the rarer medium (a)
o
to bend through 90 is called critical angle (Fig.12.9-b). When
the angle of incidence becomes larger than the critical angle,
no refraction occurs. The entire light is reflected back into the 90 o
denser medium (Fig.12.9-c). This is known as total internal Air Refracted ray
Glass
reflection of light.
Incident i
Example 12.4: Find the value of critical angle for water ray
o
(refracted angle = 90 ). The refractive index of water is 1.33
and that of air is 1. i = c
Solution: When light enters in air from water, Snell's law (b)
becomes sin r

sin i = n
or n sin i = sin r
o
n sin i = sin 90 Air No refracted ray
n sin i = 1 Glass
But n = 1.33 Incident i Reflected
ray
Therefore, ray
-1
i = sin [1/1.33] (c) i > c
-1
or = sin (0.752) = 48.8 o Fig. 12.9: Condition for total
Critical angle C = 48.8 o internal reflection
o
Therefore, critical angle of water is 48.8 .
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