Question

1) Calculate the refractive index of water if a light ray incident on a glass-
water interface at an angle of 45degree suffers a deviation of 15degree
(Refractive Index of glass=14)

I will mark you brainliest if correct...​

Answer 1

Step-by-step explanation:

Given that,

Incident angle i=45

0

Let critical angle be i

c

Now,

sini

c

=μ

Where, μ is the refractive index of glass

Now, again

r=45

0

−15

0

r=30

0

sinr

sini

=μ

μ=

sin30

0

sin45

0

μ=

2

μ=1.4

Now, the critical angle is

sini

c

=

μ

1

sini

c

=

1.4

1

i

c

=sin

−1

(0.71)

i

c

=45.2

0

Hence, the critical angle is 45.2

0

(b). given that,

Incident angle i=45

0

Now, from Snell’s formula

sinr

sini

=μ

sinr

sin45

=

2

2

×

2

1

=sinr

sinr=

2

1

r=30

0