How much is the lateral shift of the light?

Question

A light hits a 10mm thick plastic block (whose refractive index is 1.5) by 35 degrees. How much is the lateral shift of the light at the bottom of the block? Please show work.

Answer 1

You problem statement is not clear. Is the 35 degrees with respect to the normal to the surface or is it the angle with respect to the surface? I am going to assume it is the angle with respect to the normal but if it is not then it should be replaced by 55 degrees in the following:

Snell's law: ni*sin(Ai) = nr*sin(Ar)

n1 = refractive index of the initial medium

Ai = angle with respect to the normal to the surface with which the light hits the surface

nr = index of refraction of the other substance

Ar = angle with respect to the normal of the refracted light

ni = 1 for air

Ai = 35 degrees

nr = 1.5

(1)sin(35) = (1.5)sin(Ar)

sin(Ar) = sin(35)/1,5 = 0.38238

Ar = 22.48 degrees

The shift of the light, again this is not clear. Is this from the spot where the light would hit if there were no plastic block? Or is it the distance from the normal to the surface?

I will do both.

If the distance is from the normal then:

tan(Ar) = D/10mm = 0.4138

D = 4.138 mm

If there were no plastic the light would have impacted at:

tan(35) = S/10 = 0.7002

S = 7.002 mm from the normal to the surface (which is now gone)

So, if this is what is wanted, the light would be shifted 2.864 mm from where it would have impacted if there had been no plastic. And it would be shifted in the direction the light came from.

Note: you should try to be as precise as possible when stating a problem and what result or results you are looking for.