Question

In above figure, a light ray in air is incident at angle ${\theta }_1$ on a block of transparent plastic with an index of refraction of $1.56$. The dimensions indicated are $H=2.00cm$ and $W=3.00cm$. The light passes through the block to one of its sides and there undergoes reflection (inside the block) and possibly refraction (out into the air). This is the point of first reflection. The reflected light then passes through the block to another of its sides—a point of second reflection. If ${\theta }_1={40}^0$, on which side is the point of (a) first reflection and (b) second reflection? If there is refraction at the point of (c) first reflection and (d) second reflection, give the angle of refraction; if not, answer “none.” If ${\theta }_1={70}^0$, on which side is the point of (e) first reflection and (f) second reflection? If there is refraction at the point of (g) first reflection and (h) second reflection, give the angle of refraction; if not, answer “none.”

Answer 1

(a) We note that the upper-right corner is at an angle (measured from the point where the light enters, and measured relative to a normal axis established at that point the normal at that point would be horizontal in figure given in question) is at ${\tan }^{-1}\left(2/3\right)={33.7}^0$. The angle of refraction is given by
$n_{air}\sin {40}^0=1.56\sin {\theta }_2$.
which yields ${\theta }_2={24.33}^0$ if we use the common approximation $n_{air}=1.0$, and yields ${\theta }_2={24.34}^0$ if we use the more accurate value for $n_{air}$ found in Table below. The value is less than ${33.7}^0$, which means that the light goes to side 3.
(b) The ray strikes a point on side 3, which is $0.643cm$ below that upper-right corner, and then (using the fact that the angle is symmetrical upon reflection) strikes the top surface (side 2) at a point $1.42cm$ to the left of that corner. Since $1.42cm$ is certainly less than $3cm$ we have a self-consistency check to the effect that the ray does indeed strike side 2 as its second reflection (if we had gotten $3.42cm$ instead of $1.42cm$, then the situation would be quite different).
(c) The normal axes for sides 1 and 3 are both horizontal, so the angle of incidence (in the plastic) at side 3 is the same as the angle of refraction was at side 1. Thus,
$1.56\sin {24.3}^0=n_{air}\sin {\theta }_{air}\Rightarrow {\theta }_{air}={40}^0$.
(d) It strikes the top surface (side 2) at an angle (measured from the normal axis there, which in this case would be a vertical axis) of ${90}^0-{\theta }_2={66}^0$, which is much greater than the critical angle for total internal reflection (${\sin }^{-1}\left(n_{air}/1.56\right)={39.9}^0$). Therefore, no refraction occurs when the light strikes side 2.
(e) In this case, we have $n_{air}\sin {70}^0=1.56\sin {\theta }_2$,
which yields ${\theta }_2={37.04}^0$ if we use the common approximation $n_{air}=1.0$, and yields ${\theta }_2={37.05}^0$ if we use the more accurate value for $n_{air}$ found in Table below. This is greater than the ${33.7}^0$ mentioned above (regarding the upper-right corner), so the ray strikes side 2 instead of side 3.
(f) After bouncing from side 2 (at a point fairly close to that corner) it goes to side 3.
(g) When it bounced from side 2, its angle of incidence (because the normal axis for side 2 is orthogonal to that for side 1) is ${90}^0-{\theta }_2={53}^0$, which is much greater than the critical angle for total internal reflection (which, again, is ${\sin }^{-1}\left(n_{air}/1.56\right)={39.9}^0$. Therefore, no refraction occurs when the light strikes side 2.
(h) For the same reasons implicit in the calculation of part (c), the refracted ray emerges from side 3 with the same angle (${70}^0$) that it entered side 1. We see that the occurrence of an intermediate reflection (from side 2) does not alter this overall fact: light comes into the block at the same angle that it emerges with from the opposite parallel side.