Chapter 9, Problem 4C

To determine

(a)

The refractive index of the refraction for water and for fused silica.

Answer to Problem 4C

The refractive index of the refraction for water and for fused silica are $1.33$ and $1.46$ respectively.

Explanation of Solution

Given:

Velocity of light in air, $v=3\times {10}^8\text{m/s}$

Refractive index of air, $n_{air}=1$

From table 9.1, the critical angle of water, $A_2=48.6°$

The critical angle of fused silica, $A_3=43.2°$.

Formula used:

The Snell’s law is given as

$n_1\sin A_1=n_2\sin A_2$

Here,

$n_1,n_2$ are the refractive index.

$A_1,A_2$ are the critical angle.

Calculation:

The refractive index of water is calculated as

$n_1\sin A_1=n_2\sin A_2$

Plugging the values in the above equation.

$\begin{array}{c}1\sin 90°=n_{water}\sin 48.6°\\ n_{water}=1.33\end{array}$

The refractive index of fussed silica is calculated as

$n_1\sin A_1=n_3\sin A_3$

Plugging the values in the above equation.

$\begin{array}{c}1\sin 90°=n_{glass}\sin 43.2°\\ n_{glass}=1.46\end{array}$.

Conclusion:

The refractive index of the refraction for water and for fused silica are $1.33$ and $1.46$ respectively.

To determine

(b)

What is the angle of refraction?

Answer to Problem 4C

• The angle of refraction is $36.38°$.
• The angle of refraction in part (b) of the given figure is $14.7°$.
• The angle of refraction in part (c) of the given figure is $30°$.
• The angle of refraction in part (d) of the given figure is $46.9°$.
• The angle of refraction in part (e) of the given figure is $90°$.

Explanation of Solution

Given:

Refractive index of air, $n_{air}=1$

Angle of incidence, ${\theta }_i=60°$

Refractive index of glass, $n_2=1.46$

The figure of light ray.

Figure 1.

Formula used:

The Snell’s law is given as

$n_1\sin A_1=n_2\sin A_2$

Here,

$n_1,n_2$ are the refractive index

$A_1,A_2$ are the critical angle.

Calculation:

The angle of refraction is calculated as

$n_1\sin A_1=n_2\sin A_2$

Plugging the values in the above equation.

$\begin{array}{l}1\sin 60°=1.46\sin A_2\\ A_2=36.38°\end{array}$

The angle of refraction in part (b) of the figure is calculated as

$n_3\sin A_3=n_1\sin A_1$

Plugging the values in the above equation.

$\begin{array}{c}1.46\sin 10°=1\times \sin A_2\\ A_2=14.7°\end{array}$

The angle of refraction in part (c) of the figure is calculated as

$n_3\sin A_3=n_1\sin A_1$

Plugging the values in the above equation.

$\begin{array}{c}1.46\sin 20°=1\times \sin A_2\\ A_2=30°\end{array}$

The angle of refraction in part (d) of the figure is calculated as

$n_3\sin A_3=n_1\sin A_1$

Plugging the values in the above equation

$\begin{array}{c}1.46\sin 30°=1\times \sin A_2\\ A_2=46.9°\end{array}$

The angle of refraction in part (e) of the figure is calculated as

$n_3\sin A_3=n_1\sin A_1$

Plugging the values in the above equation.

$\begin{array}{c}1.46\sin 43°=1\times \sin A_2\\ A_2=90°\end{array}$.

Conclusion:

• The angle of refraction is $36.38°$.
• The angle of refraction in part (b) of the given figure is $14.7°$.
• The angle of refraction in part (c) of the given figure is $30°$.
• The angle of refraction in part (d) of the given figure is $46.9°$.
• The angle of refraction in part (e) of the given figure is $90°$