Question

The block of glass (n = 1.5) shown in cross section in Figure 1 is surrounded by air. A ray of light enters the block at its left-hand face with incident angle θ1 and reemerges into the air from the right-hand face directed parallel to the block's base. Determine the angle θ1.

Answer 1

From the figure the refraction at the second surface, by using snell's law

   

                n sin45 = n1 sin r2

               (1) sin 45 = (1.5) sinr2

then the refracting angle

            r2 = sin^-1(sin45/1.5) = 28 deg

Then from the first surface

       n sinθ1 = (1.5) sin17

         (1)   sinθ1 = 0.43855

therefore the incident angle

              θ1 = 26 deg