Deﬁnition: Snell’s Law

n

_{1}sin θ

_{1}= n

_{2}sin θ

_{}

_{2}

_{}

_{}

where

n

_{1 }= Refractive index of material 1

n

_{2 }= Refractive index of material 2

θ

_{1 }= Angle of incidence

θ

_{2 }= Angle of refraction

If

n

_{2 }<>1

then from Snell’s Law,

sin θ1 < style="font-style: italic;">θ2.

For angles smaller than 90◦, sin θ increases as θ increases. Therefore,

θ1 < θ2.

This means that the angle of incidence is less than the angle of refraction and the light ray is away toward the normal.

Similarly,if

n

_{2}> n

_{1}

then from Snell’s Law,

sin θ1 > sin θ2.

For angles smaller than 90◦, sin θ increases as θ increases. Therefore,

θ1 > θ2.

This means that the angle of incidence is greater than the angle of refraction and the light ray is bent toward the normal.

What happens to a ray that lies along the normal line?

Worked Example : Using Snell’s Law

Question:

A light ray with an angle of incidence of 35◦ passes from water to air.

Find the angle of refraction using Snell’s Law . Discuss the meaning of your answer.

(the refractive index is 1,333 for water and about 1 for air)

Answer

According to Snell’s Law: n

_{1}sin θ

_{1}= n

_{2}sin θ

_{}

_{2}1.33 sin 35◦ = 1 sin θ

_{2}sin θ

_{2}= 0.763 θ

_{2 }= 49.7◦ The light ray passes from a medium of high refractive index to one of low refractive index. Therefore, the light ray is bent away from the normal.

Test your understanding :

1. A light ray passes from water to diamond with an angle of incidence of 75◦. Calculate the angle of refraction. Discuss the meaning of your answer.

(Answer: 32.1◦, ....bent towards the normal)

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