hi everyone, i've got another question. i feel good about my methods, but i'm getting an incorrect answer.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

A ray of light impinges at a 60° angle of incidence on a glass pane of thickness 6 mm and index of refraction 1.50. The light is reflected by a mirror that touches the back of the pane (Fig. 35-37)(see attachment). By how much is the beam displaced compared with the return path it would have if the pane were absent?

2. Relevant equations

n_1/n_2 = (sin theta_2) / (sin theta_1)

opposite/adjacent= tan theta

c^2= a^2 + b^2

3. The attempt at a solution

(descriptions may be terrible w/o diagrams/letters, as I am without a drawing program just now)

so I use the first equation above to get the angle of refraction, I'll call theta_1. I've used the above tangent equation to get the bottom-length=opposite-side of the right triangle formed by the ray, the the mirror, and the vertical line to the left of the ray inside the mirror. twice this bottom length is the distance between where the ray enters the top of the glass and where the ray leaves the top of the glass, the distance between the two vertical dashed lines, a distance which I'll call x.

i need to find the distance between where the beam would 'naturally' strike the mirror and the point beneath where it leaves the glass (the right dashed line). i'll call this distance y.

To do this I need to consider a third larger right triangle formed by the left vertical dashed line, the line of the ray's 'natural' path to the mirror, and the mirror. the height of this triangle is given. its upper angle has to be 60 degrees (just like the given angle), by the geometry of two lines crossing having equal angles kitty-corner. So using the tangent formula, I can get the length of the bottom of this triangle, = the distance y, along the mirror.

Subtracting this longer distance y from the shorter distance x gives the distance between where the ray would 'naturally' strike the mirror and the point beneath where it leaves the top of the glass.

for this small right triangle, I know the height (given), and the base is y-x, so by Pythagoras, I calculate the length of the hypoteneuse.

in my understanding, the length of this hypoteneuse is the displacement of the beam and a would-be beam that never got refracted.

i tried an identical practice problem and got it right, but my answer to the homework question is wrong (somehow). any tips or suggestions would be helpful, and if there's any questions to clarify the problem, please do ask and I'll provide further details.

Again, sorry about the difficulty in description. eventually I'll figure out how to draw on OS X. (no paint brush!)

thanks much,

Yroyathon

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Physics3, angle of refraction, beam displacement

**Physics Forums | Science Articles, Homework Help, Discussion**