Chapter 23, Problem 35P

To determine

The distance between the spot of light hit at the bottom and initial point at the bottom.

Answer to Problem 35P

Solution:

The distance between the spot of light hit at the bottom and initial point at the bottom

$D=4.4\text{m}$

Explanation of Solution

Given Info:

From the given diagram,

Let $h_1=1.3\text{m}$, $h_2=2.1\text{m}$ and $d_1=2.5\text{m}$

Let PQ = $d$

Angle of incident at air-water surface be $\angle i$

Let $\angle r$ be the angle of refraction at air-water surface

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

Formula Used:

Apply Snell’s Law

$n_{air}\sin \angle i=n_{water}\sin \angle r$

Here, all alphabets are in their usual meanings.

Calculation:

First, we shall calculate incident angle

$\begin{array}{l}\cot \left(90-\angle i\right)=\frac{d_1}{h_1}\\ or,\tan \angle i=\frac{2.5\text{m}}{1.3\text{m}}\\ or,\angle i={63}^o\end{array}$

Apply Snell’s Law

$\begin{array}{l}{n}_{air}\sin \angle i=n_{water}\sin \angle r\\ or,\sin \angle r=\frac{n_{air}\sin \angle i}{n_{water}}\end{array}$

$\begin{array}{l}or,\sin \angle r=\frac{1.0\sin {63}^o}{1.33}\\ or,\angle r={42.1}^o\end{array}$

Now, we shall calculate distance ‘d’ by using triangle $\Delta TPQ$

$\begin{array}{l}\tan \angle r=\frac{PQ}{TP}\\ \tan {42.1}^o=\frac{d}{\text{2}\text{.1m}}\\ or,d=\left(\tan {42.1}^o\right).\left(\text{2}\text{.1m}\right)\\ or,d=1.\text{9}\text{m}\end{array}$

The distance between the spot of light hit at the bottom and initial point at the bottom

$\begin{array}{l}D=d_1+d\\ D=2.5\text{m+1}\text{.9 m}\\ D=4.\text{4}\text{m}\end{array}$