# Refraction!!!

• Jul 6th 2010, 05:18 PM
Cursed
Refraction!!!
White light is refracted by the triangular prism shown in the figure below. A beam of light enters the prism along a path parallel to the prism base. The light is observed on a screen that is located $\displaystyle 10m$ away from the prism and is perpendicular to the emerging rays. How far apart on the screen are the spots of blue light ($\displaystyle n=1.528$) and red light ($\displaystyle n=1.514$)

http://img208.imageshack.us/img208/1...nalproblem.png

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I'm just confused as to how you're supposed to find the distance between the spots!

I've set up the problem like so:

http://img812.imageshack.us/img812/8...ictriangle.png

$\displaystyle \theta_1= 65°$
$\displaystyle \theta_2= 180-90-65= 25°$
$\displaystyle \theta_3= 16.06°$
$\displaystyle \theta_4= 33.94°$
$\displaystyle \theta_5= 58.6°$
$\displaystyle \theta_6= 130°$

Now that I've found the angles, how do I find the distance? >.<

Work below (if you care):
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For $\displaystyle \theta_3$:

$\displaystyle n_2 sin(\theta_2) = n_3 sin(\theta_3)$
$\displaystyle (1) sin(25°)= 1.528(sin(\theta_3))$, $\displaystyle \theta_3 = 16.06°$

For $\displaystyle \theta_6$:

$\displaystyle \theta_6 + 50° + 90° + 90° = 180$, $\displaystyle \theta_6 = 130°$

For $\displaystyle \theta_4$:

$\displaystyle \theta_3 +\theta_4 +\theta_6 = 180$, $\displaystyle \theta_4=33.94°$

For $\displaystyle \theta_5$:

$\displaystyle n_4 sin(\theta_4) = n_5 sin(\theta_5)$
$\displaystyle 1.528 sin(33.94°)= (1)(sin(\theta_5))$, $\displaystyle \theta_5 =58.6°$
• Jul 6th 2010, 06:54 PM
sa-ri-ga-ma
In the problem, from which point the distance of the screen is 10 m?

The deviation of the ray from the original direction is $\displaystyle \theta_5 - \theta_2$

The deviation d of the spot on the screen is given by $\displaystyle tan(\theta_5 - \theta_2) = d/10$
• Jul 6th 2010, 07:27 PM
Cursed
I have no idea about the distance. That's partly what confused me, too. o.O

I'm assuming it's the distance from the bottom right corner to the screen, although I'm rather clueless because the problem isn't specific.

Hm. I'll try what you suggested and see how it works out. (I'm a bit tired now and am going to bed. haha.) Thanks. :)
• Jul 6th 2010, 08:05 PM
sa-ri-ga-ma
Quote:

Originally Posted by Cursed
I have no idea about the distance. That's partly what confused me, too. o.O

I'm assuming it's the distance from the bottom right corner to the screen, although I'm rather clueless because the problem isn't specific.

Hm. I'll try what you suggested and see how it works out. (I'm a bit tired now and am going to bed. haha.) Thanks. :)

Most probably the distance is measured from the point of incidence on the left face.