Question

A light ray in incident on air liquid interface at 45° and is refracted at 30°then the refrativd inndex of liquid is................. ​

Answer 1

Answer:

ANSWER

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