Chapter 23, Problem 39P

To determine

To calculate the distance from just above the spot to the strike point.

Answer to Problem 39P

Solution:

The required distance must be greater than 93.5 cm.

Explanation of Solution

Given Info:

Refractive index of water $n_{water}=1.33$

Refractive index of air $n_{air}=1.0$

Water depth ‘D’ = 82.0 cm

Let the angle of incident (critical angle)be $\angle i=\angle c$

Let distance from just above the spot to the strike point be ‘d’

Angle of refraction $\angle r={90}^0$

Formula Used:

Apply Snell’s Law

$\begin{array}{l}{n}_{water}\sin \angle i=n_{air}\sin \angle r\\ n_{water}\sin \angle c=n_{air}\sin {90}^0\end{array}$

Here, all alphabets are in their usual meanings.

Calculation:

Apply Snell’s Law for water-air surface

$\begin{array}{l}{n}_{water}\sin \angle i=n_{air}\sin \angle r\\ n_{water}\sin \angle c=n_{air}\sin {90}^0\\ or,\sin \angle c=\frac{n_{air}\sin {90}^0}{n_{water}}\\ or,\sin \angle c=\frac{1.0\sin {90}^0}{1.33}\\ or,\angle c={48.7535}^0\end{array}$

Now, we shall calculate the distance ‘d’

$\begin{array}{l}\tan \angle c=\frac{d}{D}\\ or,d=D\tan \angle c\\ or,d=\left(82.0cm\right)\tan {48.7535}^0\\ or,d=93.5cm\end{array}$

Since the light should not exit from water, then angle of incident must be greater than to its critical angle. Hence, the required distance must be greater than 93.5 cm.