Question

A ray of light is incident on an interface at an angle of 450 and is refracted at

300. Find

(a) Refractive index of the rarer medium

(b) Refractive index of the denser medium​

Answer 1

Explanation:

Given : i=45

0

r=30

0

(μ

air

)

To find : μ

water

Solution: From snell's law

μ

1

sini=μ

2

sinr

μ

air

sin45

0

=μ

w

sin30

0

2

1

=

2

μ

w

μ

w

2

Now let us take that for angle of incidence α the angle between reflected ray and refracted ray be 90

0

then

α+r+90

0

=180

0

/90

0

(since i = angle of reflection from laws of reflection)

Hence r=90−α

Now from snell's law

μ

1

sinα=μ

2

sinr

1sinα=

2

sin(90−α)

tanα=

2

α=tan

−1

2