Question

A ray of light traveling from air enters a liquid at an angle of 45 degree with the normal. if the corresponding angle of refraction is 30 degree then the refractive index of the liquid with respect to air is​

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

Answer 2

Given:-

• A ray of light traveling from air enters a liquid at an angle of 45° with the normal.
• The corresponding angle of refraction is 30°.

⠀

To find:-

• The refractive index of the liquid with respect to air.

⠀

Solution:-

★ In this question we have given that a ray of light traveling from air enters a liquid at an angle of 45° with the normal. The corresponding angle of refraction is 30°. We have to find out the refractive index of the liquid with respect to air. Let's do it.

⠀

According to the question,

⠀

⇢ ix sin45° = rx sin30°

⇢ 1/√2 = r × 1/2

⇢ 2/√2 = r

⇢ r = √2

⠀

Hence,

• the refraction index of the liquid with respect to air is √2.