Question

Consider a light wave that passes from air into a very thick clear glass block that has its opposite internal side mirrored (facing into the glass). The light ray passes into the glass block at an angle greater than $0^{\circ }$ and less than ${90}^{\circ }$, strikes the mirrored surface and reflects back through the glass into the air.
What happens when the light ray leaves the glass block after it has struck the mirrored surface?

• A. The waves reflect back on the glass
• B. The waves increase velocity
• C. The waves leave the glass at the same angle at which they entered the glass
• D. The angle of refraction is greater than the angle of incidence
• E. The angle of reflection from the mirror is equal to the incident angle at which the light struck the glass block

Answer 1

The correct option is C The waves leave the glass at the same angle at which they entered the glass

The angle of reflection from mirror is equal to the angle of incidence on the mirror. Since light has refracted at the glass-air interface, incidence angles at the two surfaces cannot be the same.

However the ray leaving the glass follows snell's law,

${\mu }_{g}sin{r}_2={\mu }_{a}sinr$

The light entering glass followed,

${\mu }_{a}sini={\mu }_{g}sin{r}_1$

Since $r_1=r_2$, we have

$i=r$

Hence correct answer is option C.