Chapter 23, Problem 33P

To determine

Part (a) To Determine:

The angle of refraction (a)At air-glass surface.

Answer to Problem 33P

Solution:

The angle of refraction

At air-glass surface $\angle r_{a-g}={26.6}^o$

Explanation of Solution

Given Info:

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

Refractive index of glass $n_{glass}=1.54$

Angle of incident at air-glass surface $\angle i_{a-g}={43.5}^o=\angle i_{a-w}$

Let the angle of refraction at air-glass surface be $\angle r_{a-g}$

Formula Used:

Apply Snell’s Law

$n_1\sin \angle i=n_2\sin \angle r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply Snell’s Law at air-glass surface

$\begin{array}{l}{n}_{air}\sin \angle i_{a-g}=n_{glass}\sin \angle r_{a-g}\\ or,\sin \angle r_{a-g}=\frac{n_{air}\sin \angle i_{a-g}}{n_{glass}}\\ or,\sin \angle r_{a-g}=\frac{1.0\sin {43.5}^o}{1.54}\\ or,\angle r_{a-g}={26.6}^o\end{array}$

To determine

Part (b) To Determine:

To calculate the angle of refraction (b)At glass-water surface.

Answer to Problem 33P

Solution:

The angle of refraction (b)At glass-water surface $\angle r_{g-w}={34.6}^o$

Explanation of Solution

Given Info:

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

Refractive index of glass $n_{glass}=1.54$

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

Angle of incident at air-glass surface $\angle i_{a-g}={43.5}^o=\angle i_{a-w}$

Angle of incident at glass-water surface $\angle i_{g-w}=\angle r_{a-g}$

Let the angle of refraction at glass-water surface be $\angle r_{g-w}$

Formula Used:

Apply Snell’s Law

$n_1\sin \angle i=n_2\sin \angle r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply Snell’s Law at glass-water surface

$\begin{array}{l}{n}_{glass}\sin \angle i_{g-w}=n_{water}\sin \angle r_{g-w}\\ or,\sin \angle r_{g-w}=\frac{n_{glass}\sin \angle i_{g-w}}{n_{water}}\\ or,\sin \angle r_{g-w}=\frac{1.54\sin {26.6}^o}{1.33}\\ or,\angle r_{g-w}={34.6}^o\end{array}$

To determine

Part (c) To Determine:

The angle of refraction (c)At air-water surface (when there is no glass)

Answer to Problem 33P

Solution:

The angle of refraction (c)At air-water surface (when there is no glass)$\angle r_{a-w}={31.2}^o$

Explanation of Solution

Given Info:

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

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

Let the angle of refraction at air-glass surface be $\angle r_{a-g}$

Angle of incident at glass-water surface $\angle i_{g-w}=\angle r_{a-g}$

Let the angle of refraction at air-water surface be $\angle r_{a-w}$

Formula Used:

Apply Snell’s Law

$n_1\sin \angle i=n_2\sin \angle r$

Here, all alphabets are in their usual meanings.

Calculation:

Apply Snell’s Law at air-water surface

$\begin{array}{l}{n}_{air}\sin \angle i_{a-w}=n_{water}\sin \angle r_{a-w}\\ or,\sin \angle r_{a-w}=\frac{n_{air}\sin \angle i_{a-w}}{n_{water}}\\ or,\sin \angle r_{a-w}=\frac{1.0\sin {43.5}^o}{1.33}\\ or,\angle r_{a-w}={31.2}^o\end{array}$