When the light ray illustrated in the figure below passes through the glass bloc

Question

When the light ray illustrated in the figure below passes through the glass block of index of refraction n = 1.50, it is shifted laterally by the distance d. (Let L = 2.78 cm and ? = 27.0

When the light ray illustrated in the figure below passes through the glass block of index of refraction n = 1.50, it is shifted laterally by the distance d. (Let L = 2.78 cm and ? = 27.0 A degree.) (a) Find the value of d. cm (b) Find the time interval required for the light to pass through the glass block. ps

Explanation / Answer

From Snell's Law:
[sin(27deg)]/[sin(r)] =1.5/1
= (index of refraction of the glass)/(index of refraction of air)

sin(r) = [sin(27deg)]/(1.5) = (0.453)/(3/2) = 0.680
So
r = arcsin(0.680) = 42.90 deg
Construct an extension of the air-side normal to the glass all the way straight through the glass, and you will be able to see that the ray's path splits the 27 degree angle inside the glass into the angle A and another part that is (27-r) degrees.
Thus you can set up the equations:
(2cm)/L = cos(r) and
d/L = sin(27-r) ;
where L is the ray's path length through the glass, and it forms the hypotenuse of both triangles involved in the two equations. You can then solve the first equation for the length of L and substitute that into the second equation to find the length of d

d= 0.273 cm

2) Use the Snells Law equations (url below). Start with n(air) and v(air) that you should know, and use n(glass)=1.5 and to get v(glass). Likewise, use n(air), theta(air), and n(glass) to get theta(glass).

v= d/t general def of speed.

here t= d/v             we know that n= c/v so v=c/n substitute in t, t= d.n/c

   Time t= 2*10^-2*1.5/3*10^8m

            = 1 *10^-10 s

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