Solving Light Optics Problem: Find Reflection Point

[jena]

Hi,

My Question:

A beam of light is emitted in a pool from a depth of 62.0 cm. Where must it strike the air-water interface, relative to the spot directy abovie, in order that the light does not exit the water?


My Work

Tan(49)=(x/62.0cm), got 49 degrees from the textbook where it says that it won't reflect at this angle.

x=Tan(49)62.0 cm

x= 71.3 cm

Is this right??

Thank You

 

[mrjeffy321]

First you are looking for the "critical angle" a which the light will be experience total internal reflection at the water-air interface.
The critical angle is equal to the arcsin (index of refraction of air / index of refraction of water).
the index of refraction of air is about 1, and water is about 1.33,
so you can find the critical angle is just about 49 degrees like you book said.

If the light is at an angle of 49 degrees, and it travels up to the water-air interface a distance of 62 cm then what would the x distance be?
We know that R*sin(49) = 62, so R is around 82.15 cm, and we know that,
R*cos(49) = the x distance traveled by the light.
I get something different that your answer when I plug in the values and sove for x.

 

[verty]

The answer is correct for 49 degrees. For 48.6 it is 70.3cm.

 

[jena]

Hi,

I just found another way but it still uses sine and also finds how I came up with the 49 degrees.

What I did was first find the critical angle for the light leaving the water

n(water)*sin(theta)=n(air)sin(theta)

1.33sin(theta)=1*sin(90) where I found theta to be 48.8

I used it in the equation where Tan(48.8)=(x/62.0cm), and the answer comes to about 70.7cm.

Could this work

 

[verty]

That's the proper method, but how did you get 48.8 degrees?

 

[verty]

Ok, I see what it is, 1.33 gives 48.8, 1.333 gives 48.6. Certainly it seems to be given as 1.33. When I did this in school the angle was given as 48.6, so that's why I wondered about the angle. I don't know which is correct, but rather use 1.33 unless the teacher says otherwise.