Chapter 18, Problem 46A

a.

To determine

Angle made by beam while entering the glass.

Answer to Problem 46A

  ${\theta }_{\text{glass}}=25.4°$

Explanation of Solution

Given:

Refractive index of glass, $n_{\text{glass}}=1.5$

Angle of incidence of light in air is ${\theta }_{\text{air}}=40°$

Formula Used:

Snell’s law is given as,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, $n_1$ and $n_2$ is the refractive index of the two medium and ${\theta }_1$ and ${\theta }_2$ is the angle with the normal.

Calculation:

As, $n_{\text{glass}}=1.5$ and $n_{\text{air}}=1$

For air and glass medium the Snell’s law is

  $\begin{array}{l}{n}_{\text{air}}\sin {\theta }_{\text{air}}=n_{\text{glass}}\sin {\theta }_{\text{glass}}\\ {\theta }_{\text{glass}}={\sin }^{-1}\left(\frac{n_{\text{air}}\sin {\theta }_{\text{air}}}{n_{\text{glass}}}\right)\end{array}$

On substituting the values,

  $\begin{array}{c}{\theta }_{\text{glass}}={\sin }^{-1}\left(\frac{1\times \sin 40°}{1.5}\right)\\ =25.4°\end{array}$

Conclusion:

Therefore, the beam enters the glass at an angle ${\theta }_{\text{glass}}=25.4°$ .

b.

To determine

Angle made by beam while entering the water.

Answer to Problem 46A

  ${\theta }_{\text{water}}=28.9°$

Explanation of Solution

Given:

Refractive index of glass, $n_{\text{glass}}=1.5$

Angle of incidence of beam in air is ${\theta }_{\text{air}}=40°$

Formula Used:

Snell’s law is given as,

  $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, $n_1$ and $n_2$ is the refractive index of the two medium and ${\theta }_1$ and ${\theta }_2$ is the angle with the normal.

Calculation:

From previous subpart, ${\theta }_{\text{glass}}=25.4°$

Now for glass and water medium the Snell’s law is

  $\begin{array}{l}{n}_{\text{glass}}\sin {\theta }_{\text{glass}}=n_{\text{water}}\sin {\theta }_{\text{water}}\\ {\theta }_{\text{water}}={\sin }^{-1}\left(\frac{n_{\text{glass}}\sin {\theta }_{\text{glass}}}{n_{\text{water}}}\right)\end{array}$

As, $n_{\text{water}}=1.33$ and $n_{\text{glass}}=1.5$

On substituting the values,

  $\begin{array}{c}{\theta }_{\text{water}}={\sin }^{-1}\left(\frac{1.5\times \sin 25.4°}{1.33}\right)\\ =28.9°\end{array}$

Conclusion:

Therefore, the beam enters the water at an angle ${\theta }_{\text{water}}=28.9°$ .