Is Snell's Law valid even when incident ray is Normal to the surface?

[aleemudasir]

According to the Snell's Law refractive index n21= sin i/sin r, but when we use this equation while having a incident light normal to the surface of lens or any other refracting surface it becomes 0/0. So how can we define Snell's law in this situation?

 

[Infinitum]

$$n_1sin\theta_1 = n_2sin\theta_2$$

This is the actual Snell's law. When you shift the sin over to the other side, you assume it's not 0.

 

[aleemudasir]

$$n_1sin\theta_1 = n_2sin\theta_2$$

This is the actual Snell's law. When you shift the sin over to the other side, you assume it's not 0.

Which one is the actual n21=sin i /sin r or n1 sin i= n2 sin r?

 

[Infinitum]

As I said above...

$$n_1sin(i) = n_2sin(r)$$

The other one is not valid when sin(r) is equal to zero.

 

[sardar]

Yes, Snell's Law is still valid even when the incident ray is normal to the surface. In this case, the angle of incidence (i) is equal to 0 degrees and the refracted ray will also be normal to the surface, resulting in an angle of refraction (r) of 0 degrees. This means that the ratio of sine of the angles (n21) will also be equal to 0/0, which is undefined. However, this does not invalidate the law because the refractive index (n) is not dependent on the angle of incidence or refraction. It is a constant value for a specific material. Therefore, in this situation, Snell's Law can still be defined as n1sin0 = n2sin0, which simplifies to n1 = n2. This means that the refractive index of the material on both sides of the interface is the same, resulting in no refraction.