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Find your uni forum to get talking to other applicants, existing students and your future course-mates 27-07-2015
1. Once again I need help from you geniuses out there....

Question:

"A Light beam is incident on a plane sheet of glass (thickness 5mm) at an angle of incidence of 30 degrees. What is the lateral displacement, t, of the transmitted light relative to that followed by the beam with no glass present?"

I have no idea where to start from, hopefully someone can help me out.

Thanks.
2. Draw a diagram...... two parallel lines to represent the glass.
Draw the normal in at the first air-glass boundary and label the angle 30. Label this boundary crossing '1'

(Using refractive index of glass = 1.3 here....)

Take the eqn "sin theta1 over sin theta2 = n2 over n1"
work this bad boy out an you get theta2=22.6 degrees
make a line at about 20 degrees on the diagram through the glass (only needs to represent)

So the ray emerges at the same angle as it originally hit.
on the air-glass boundary of the first bit, the angle is 22.6
the distance across the glass (extension of the normal line) is 5mm
using trigonometry, we can find the distance "opposite" the 22.6 angle, where the "adjacent" side is the depth of the glass (known) and the "hypotenuse" is the ray passing through it.

we know the "adjacent" and "opposite" values so tan 22.6 = o/a
therefore the "opposite" side is equal to ["adjacent" tan 22.6]

sorry if this is hard to get your head round - writing out equations is hard on forums, but this "opposite" length is the offset between the incident ray entering the glass and the resultant ray leaving the glass, which is hw i've interpreted the question.

hope this helps.
3. (Original post by corkskrew)
Draw a diagram...... two parallel lines to represent the glass.
Draw the normal in at the first air-glass boundary and label the angle 30. Label this boundary crossing '1'

(Using refractive index of glass = 1.3 here....)

Take the eqn "sin theta1 over sin theta2 = n2 over n1"
work this bad boy out an you get theta2=22.6 degrees
make a line at about 20 degrees on the diagram through the glass (only needs to represent)

So the ray emerges at the same angle as it originally hit.
on the air-glass boundary of the first bit, the angle is 22.6
the distance across the glass (extension of the normal line) is 5mm
using trigonometry, we can find the distance "opposite" the 22.6 angle, where the "adjacent" side is the depth of the glass (known) and the "hypotenuse" is the ray passing through it.

we know the "adjacent" and "opposite" values so tan 22.6 = o/a
therefore the "opposite" side is equal to ["adjacent" tan 22.6]

sorry if this is hard to get your head round - writing out equations is hard on forums, but this "opposite" length is the offset between the incident ray entering the glass and the resultant ray leaving the glass, which is hw i've interpreted the question.

hope this helps.

Thanks, that helped me alot. Im just wondering that you automatically said that the ray tha emerges out at the same angle it enters the glass which was 30degrees. How can u make this assumption???,....and also was theta2 the angle of refraction?
4. Decided to supplement with a diagram! (attached)
Attached Thumbnails

5. yes, theta 2 is the angle of refraction.
the emergant ray is always parallel to the incident ray provided the two faces through which it pass are parallel - you can work the angles out using the refraction law if you like, but it is the same. if you do the equation again, you are literally just reversing the whole thing for the air-glass crossing
6. (Original post by corkskrew)
yes, theta 2 is the angle of refraction.
the emergant ray is always parallel to the incident ray provided the two faces through which it pass are parallel - you can work the angles out using the refraction law if you like, but it is the same. if you do the equation again, you are literally just reversing the whole thing for the air-glass crossing
thanks....hmmm..
the diagram u hav drawn has puzzled me abit cos i thought that the thickness which is 5mm was the adjacent side but in the diagram u have drawn it in the opposite side....we want to know the lateral displacement, but the diagram labels the lateral displacement as the thickness???
7. blue horizontal lines in the diagram are the glass edges.
therefore the thickness = "adj" = 5mm
and also the displacement = "s" or "opp" = 5 tan theta2 (in mm)
8. the thick black lines represent the normal lines (at 90 degrees to the surface - the angle between this and the incident ray gives the angle of incidence)
Updated: June 2, 2005
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