# Math Help - Refraction!!!

1. ## 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 $10m$ away from the prism and is perpendicular to the emerging rays. How far apart on the screen are the spots of blue light ( $n=1.528$) and red light ( $n=1.514$)

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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:

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

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

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

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

For $\theta_6$:

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

For $\theta_4$:

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

For $\theta_5$:

$n_4 sin(\theta_4) = n_5 sin(\theta_5)$
$1.528 sin(33.94°)= (1)(sin(\theta_5))$, $\theta_5 =58.6°$

2. In the problem, from which point the distance of the screen is 10 m?

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

The deviation d of the spot on the screen is given by $tan(\theta_5 - \theta_2) = d/10$

3. 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.

4. 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.