Question

- Fig. (a) and (b) show refraction of an incident ray in air at 60° with the normal to a glass-air and
water-air interface, respectively. Predict the angle (r) of refraction of an incident ray in water at
45° with the normal to a water-glass interface (Fig. (c)].
60!
V GLASS
Y
AIR
7
GLASS
AIR
-WATER
TER:
60°7
----745°5=====
-
-----
(0)
(b)
(C)​

Answer 1

Answer:

question

Asked on November 22, 2019 by

avatar

Jain Tulasi

1 Answer

Refractive index of glass \mu_g=sin60^o/sin35^o=1.51μ

g

=sin60

o

/sin35

o

=1.51

For water \mu_w=sin60^o/sin47^o=1.184μ

w

=sin60

o

/sin47

o

=1.184

So, refractive index of glass wrt to water can be \mu_r=\mu_g/\mu_w=1.275μ

r

=μ

g

/μ

w

=1.275

now,

\mu_r=sin45^o/sinr=1.275\Rightarrow r=36.68^o\approx 37^o