Chapter 33 - Electromagnetic Waves - Problems - Page 1006: 66d

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Let $\theta_2$ be the refracted angle inside the block. We can use Snell's law to find $\theta_2$: $n_2~sin~\theta_2 = n_1~sin~\theta_1$ $sin~\theta_2 = \frac{n_1~sin~\theta_1}{n_2}$ $\theta_2 = sin^{-1}~(\frac{n_1~sin~\theta_1}{n_2})$ $\theta_2 = sin^{-1}~(\frac{1.00~sin~40^{\circ}}{1.56})$ $\theta_2 = sin^{-1}~(0.41204)$ $\theta_2 = 24.3^{\circ}$ If the ray travels a horizontal distance of $W = 3.00~cm$, we can find the vertical distance $y$ that the ray travels: $\frac{y}{W} = tan~\theta_2$ $y = W~tan~\theta_2$ $y = (3.00~cm)~tan~24.3^{\circ}$ $y = 1.36~cm$ Since $y \lt H$, then the point of first reflection is on side 3. The reflected angle is equal to the incident angle on side 3. Since $2y \gt H$, then the reflected ray meets side 2 before it is able to cross the horizontal distance of $W = 3.00~cm$ back to side 1. The point of second reflection is on side 2. Let $\theta_4$ be the refracted angle in the air. Note that the incident angle at the point of second reflection is $90^{\circ}-24.3^{\circ}$ which is $\theta_i = 65.7^{\circ}$ We can use Snell's law to find $\theta_4$: $n_4~sin~\theta_4 = n_2~sin~\theta_i$ $sin~\theta_4 = \frac{n_2~sin~\theta_i}{n_4}$ $sin~\theta_4 = \frac{1.56~sin~65.7^{\circ}}{1.00}$ $sin~\theta_4 = 1.42$ Since there is no angle such that $sin~\theta_4 = 1.42$, there is no refraction at the point of second reflection.