a room which is h = 9.0 feet tall and r = 12.0 feet wide. Attached to the ceilin

Question

a room which is h = 9.0 feet tall and r = 12.0 feet wide. Attached to the ceiling is a piece of glass (with unknown index of refraction) which is 2.5 feet thick. A laser pointer in the bottom left corner is aimed so that the ray which reflects off the glass hits the room's bottom right corner.


(a) What is the ray's angle of incidence at the glass?
°
(b) If the ray which refracts into the glass hits the ceiling 7.7 feet away from the left wall, what is the refracted ray's angle of refraction at the glass?
°
(c) What is the glass's index of refraction?

Explanation / Answer

If the ray hits at the bottom left corner of the room, this means that it hits the glass right at the midpoint.
Remember SOH CAH TOA

In this case we will use Tan to find the angle the the ray forms with the bottom left corner of the room:
to find the midpoint: 12/2=6ft
to find the height: 9-2.5ft(from the glass)=6.5ft thus,

theta=tan-1(6.5/6)=26.6 then 90-26.6=63.4 degrees A) 63.4 degrees

B) The ray's angle of refraction hits the ceiling of the room at 7.7 thus:

12-7.7=4.3ft

The angle of refraction creates a triangle with the normal line which is 5ft long.

4.3 is the opposite side of the angle of refraction, and 5 is the adjacent side.

So again we use Tan.

theta=tan-1(4.3/50)=40.7degrees

C) To find the index of refraction use Snell's Law

n1sin(theta1)=n2sin(theta2)

*keep in mind that the angle of refraction is less than the angle of incidence so the refracted angle needs to go towards the normal line.

n1=1

theta1: 63.4

n2=?

theta2=40.7

sin(63.4)=n2sin(40.7)

sin(63.4)/sin(40.7)=n2

n2=1.37

A)63.4 degrees

B) 40.7 degrees

C) 1.37

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