# Maximum angle of the refracted light beam

• Tardis Traveller
In summary, the problem involves a light ray entering a cylindrical glass tube with an index of refraction of 3/2 and submerged in water with an index of refraction of 4/3. The maximum entry angle at which the light ray will only travel inside the glass can be found using Snell's law, and will result in the light ray being reflected within the glass at a critical angle.

## Homework Statement

A light ray falls from the air (##n_a=1##) into the center of the upper surface of a long cilindrical glass tube with an index of refraction ##n_t=3/2##. The tube is submerged into water all the way to the upper edge and the waters index of refraction is ##n_w=4/3##. What is the maximum entry angle of the light ray at which it will only be traveling inside the glass?
a)##arcsine(1/6)##
b)##arcsin(\sqrt{15}/6)##
c)##arcsin(\sqrt{17}/6)##

## Homework Equations

3. The Attempt at a Solution
I gues the snells law must be used ##n_1sin(x_1)=n_2sin(x_2)## but i don't get the problem. What is meant by maximum entry angle at which the beam will be traveling inside the cylinder only and how do i get that? I don't know my starting points. Could you hint?[/B]

Tardis Traveller said:
dont get the problem
The light ray strikes the top of the cylinder at some angle. Refraction rotates it to a steeper angle within the glass. What might happen next?

haruspex said:
The light ray strikes the top of the cylinder at some angle. Refraction rotates it to a steeper angle within the glass. What might happen next?
I thought about it and i think the refracted ray inside the glass should come to the edge at a critical angle so that all of it gets reflected to the glass again and so on right?

Tardis Traveller said:
I thought about it and i think the refracted ray inside the glass should come to the edge at a critical angle so that all of it gets reflected to the glass again and so on right?
Yes, though I'm not sure what you mean by "and so on" there.