Question

A ray of light strikes a flat block of glass $$ (n=1.50) $$ of thickness $$ 2.00 \mathrm{cm} $$ at an angle of $$ 30.0^{\circ} $$ with the normal. Trace the light beam through the glass and find the angles of incidence and refraction at each surface.

Answer 1

At entry, the wave under refraction model, expressed as $$ n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2}, $$ gives

$$\theta_{2}=\sin ^{-1}\left(\dfrac{n_{1} \sin \theta_{1}}{n_{2}}\right)=\sin ^{-1}\left(\dfrac{1.000 \sin 30.0^{\circ}}{1.50}\right)=19.5^{\circ}$$

To do ray optics, you must remember some geometry. The surfaces of entry and exit are parallel so their normals are parallel. Then angle $$ \theta_{2} $$ of refraction at entry and the angle $$ \theta_{3} $$ of incidence at exit are alternate interior angles formed by the ray as a transversal cutting parallel lines. Therefore, $$ \theta_{3}=\theta_{2}=\left[19.5^{\circ}\right] $$

At the exit point, $$ n_{2} \sin \theta_{3}=n_{1} \sin \theta_{4} $$ gives

$$\theta_{4}=\sin ^{-1}\left(\dfrac{n_{2} \sin \theta_{3}}{n_{1}}\right)=\sin ^{-1}\left(\dfrac{1.50 \sin 19.5^{\circ}}{1.000}\right)=30.0^{\circ}$$

Because $$ \theta_{1} $$ and $$ \theta_{4} $$ are equal, the departing ray in air is parallel to the original ray.