Calculating Angle of Refraction for Light Passing Through Water and Glass

[aliaze1]

Homework Statement

A 1.0-cm-thick layer of water stands on a horizontal slab of glass. A light ray in the air is incident on the water 60 degrees from the normal.

What is the ray's direction of travel in the glass?

Homework Equations

n΄sinθ΄ = n¹sinθ¹

or, if that is confusing,

n1sinθ1 = n2sinθ2

The Attempt at a Solution

The index of refraction for water is 1.33, and the index of refraction for air is 1

therefore i set up the problem as:

(1)sin(60) = (1.33)sinθ

θ should = about 40.628 degrees, but the computer says it is incorrect... where am I going wrong?

thanks

 

[learningphysics]

What exactly does the question ask for?

 

[aliaze1]

oh sorry about that, it asks this

'What is the ray's direction of travel in the glass?'

i just edited it also

 

[learningphysics]

oh sorry about that, it asks this

'What is the ray's direction of travel in the glass?'

i just edited it also

So you did the first part to get the angle across the air/water interface... but now you need to handle the water/glass interface... actually I'm thinking since:

n1sin(theta1) = n2sin(theta2)... n2sin(theta2) = n3sin(theta3)...

You can simply ignore the water part...

n1sin(theta1) = n3sin(theta3)

unless I'm missing something...

 

[aliaze1]

Lol!

So you did the first part to get the angle across the air/water interface... but now you need to handle the water/glass interface... actually I'm thinking since:

n1sin(theta1) = n2sin(theta2)... n2sin(theta2) = n3sin(theta3)...

You can simply ignore the water part...

n1sin(theta1) = n3sin(theta3)

unless I'm missing something...

LOL oh i see! hahaha i was thinking that the water itself was the reflective surface and ignored the word 'glass'! haha this should work

 

[aliaze1]

yay!

yes you were correct, it works!

using the other index as 1.5 (ignoring the water entirely), i get 1sin(60)=1.5sin(x), and then i just had to solver for x, which gave me the correct value of ~35.3 deg

thanks!

 

[learningphysics]

yes you were correct, it works!

using the other index as 1.5 (ignoring the water entirely), i get 1sin(60)=1.5sin(x), and then i just had to solver for x, which gave me the correct value of ~35.3 deg

thanks!

no prob! you're welcome!