Chapter 22, Problem 40P

(a)

To determine

The minimum incident angle would ray of light undergo total internal reflection.

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection is $34.2°$.

Explanation of Solution

The ray diagram for the minimum incident angle is given below:

Formula to calculate the minimum incident angle is,

${\theta }_i={\sin }^{-1}\left(\frac{n_a\sin \left(90\right)}{n_g}\right)$

Here,

${\theta }_i$ is the minimum incident angle

$n_a\text{and}{n}_g$ are the refractive index of air and glass

Substitute $1$ for $n_a$, $1.78$ for $n_g$ to find ${\theta }_i$,

$\begin{array}{c}{\theta }_i={\sin }^{-1}\left(\frac{\left(1\right)\sin \left(90\right)}{\left(1.78\right)}\right)\\ =34.2°\end{array}$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$.

(b)

To determine

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass.

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass is $34.2°$.

Explanation of Solution

The ray diagram for the minimum incident angle for the glass water interface is,

Expressing the formula for the critical angle is,

$\begin{array}{l}{n}_w\sin {\theta }_c=n_a\sin \left(90\right)\\ \sin {\theta }_c=\frac{1}{n_w}\end{array}$

Here,

${\theta }_c$ is the critical angle

$n_w$ is the refractive index of water

$n_a$ is the refractive index of air

Formula to calculate the minimum incident angle is,

$\begin{array}{c}{\theta }_i={\sin }^{-1}\left(\frac{n_w\sin {\theta }_c}{n_g}\right)\\ ={\sin }^{-1}\left(\frac{n_w\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n_w$}\right.\right)}{n_g}\right)\\ ={\sin }^{-1}\left(\frac{1}{n_g}\right)\end{array}$

Here,

${\theta }_i$ is the minimum incident angle

$n_g$ is the refractive index of glass

Substitute, $1.78$ for $n_g$ to find ${\theta }_i$,

$\begin{array}{c}{\theta }_i={\sin }^{-1}\left(\frac{\left(1\right)}{\left(1.78\right)}\right)\\ =34.2°\end{array}$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$.

(c)

To determine

The effect of the thickness of water layer or glass on the minimum incident angle.

Answer to Problem 40P

The thickness of water layer or glass does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

${\theta }_i={\sin }^{-1}\left(\frac{1}{n_g}\right)$

Here,

${\theta }_i$ is the minimum incident angle

$n_g$ is the refractive index of glass

Thus, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

(d)

To determine

The effect of the refractive index of the intervening layer on the minimum incident angle.

Answer to Problem 40P

The refractive index of the intervening layer does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

${\theta }_i={\sin }^{-1}\left(\frac{1}{n_g}\right)$

Here,

${\theta }_i$ is the minimum incident angle

$n_g$ is the refractive index of glass

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the refractive index of the intervening layer.