Total internal reflection: Question on angle of reflection.

[ssd]

Homework Statement

Suppose light from is moving from water to air. The critical angle be θ i.e. for angle of incidence θ, the angle of refraction is 90°. Refractive index of water wrt air be μ.
Suppose the angle of incidence (i) is increased by 1° to θ+1. What happens to the refracted/reflected ray. The angle of refraction (r) would be (?) given by sin r= μ*sin(θ+1). That is if μ =1.33, then θ≈49° say. Then for i= 50°, sin r>1... not solvable by Snail's law.
Due to total internal reflection, r = 90+ (90-50)= 130°.

My confusion is, if i is increased from θ by a small amount, will not r also be increased from 90° by a small amount? But due to total internal reflection r is increased by a big amount like 40° !

Homework Equations

The Attempt at a Solution

I am not a student. This question is asked by my daughter who is 14 yr old. And I am confused. Please suggest the solution . Thanks.

 

[sankalpmittal]

Homework Statement

Suppose light from is moving from water to air. The critical angle be θ i.e. for angle of incidence θ, the angle of refraction is 90°. Refractive index of water wrt air be μ.
Suppose the angle of incidence (i) is increased by 1° to θ+1. What happens to the refracted/reflected ray. The angle of refraction (r) would be (?) given by sin r= μ*sin(θ+1). That is if μ =1.33, then θ≈49° say. Then for i= 50°, sin r>1... not solvable by Snail's law.
Due to total internal reflection, r = 90+ (90-50)= 130°.

My confusion is, if i is increased from θ by a small amount, will not r also be increased from 90° by a small amount? But due to total internal reflection r is increased by a big amount like 40° !

Homework Equations

The Attempt at a Solution

I am not a student. This question is asked by my daughter who is 14 yr old. And I am confused. Please suggest the solution . Thanks.

No matter how much you increment the angle of incidence from critical angle, total internal reflection will take place if light ray is going from denser to rarer medium. Hence laws of reflection comes into play and Snell's law fails.

 

[CWatters]

I haven't followed your description but perhaps see youtube demo..



Yes. At the critical angle a small change in the input angle results in big change in the angle of the emerging ray.

In some situations (combinations of object and observer) this results in a "blind spot".

(Edited for typos).

 

[CWatters]

This one shows it slightly more clearly..



Note that close to the critical angle you can get both refraction and reflection. The input beam is split.