Largest incident angle a ligth beam can have and still exit the glass?
A beam of light rays strike a square glass plate of thickness .80 cm at an angle of incidence of 60 degrees. How much will the beam displace from its original line of travel? What is the largest incident angle a light beam can have and still exit the glass?
Stare
on February 8th, 2012
When the beam enters the glass, its angle with respect to the normal will be reduced: by Snell’s law,
sin(A) = sin(60 degrees)/(index_of_refraction)
= 0.866/n
where A = new angle
The distance parallel to the glass surface that the ray would have traveled, without deflection, would have been:
d = tan(60 degrees) * thickness
= 1.732 * 0.80 cm = 1.386 cm
The distance parallel to the glass surface that the deflected ray travels is instead:
d’ = tan(A) * thickness
= tan(A) * 0.80 cm
tan(A) = sin(A)/cos(A) = sin(A)/sqrt(1 – (sin(A))^2)
= (0.866/n)/sqrt(1 – ((0.866/n)^2))
= 0.866/sqrt(n^2 – (0.866)^2)
Therefore, the displacement between these two paths is:
d – d’ = 1.386 – 0.693/sqrt(n^2 – 0.75)
Unfortunately, you did not provide the index of refraction of this piece of glass! I will arbitrarily select a value of n = 1.51.
Then the displacement is:
d – d’ = 1.386 – 0.693/sqrt(1.51^2 – 0.75)
= 0.826 cm
Now, with regards to the question of largest incident angle: Any incidence angle is OK, because as long as the two sides of the glass pane are parallel, if the ray can get into the pane, it can get out. The deflection of the angle coming in will just be reversed upon exiting. And for light entering a medium of higher refractive index, there is no limitation on the angle of incidence. So any angle up to 90 degrees is OK.