Chapter 35, Problem 66AP

To determine

The reason for the situation to be impossible.

Answer to Problem 66AP

The given situation is impossible as the total number of reflection possible is $82$, but it is given that $85$ reflection has occurred.

Explanation of Solution

The following figure shows the entire path of the laser beam traveled through the slab.

Figure-(1)

Write the expression for critical angle.

    $\begin{array}{c}\sin {\theta }_{\text{c}}=\frac{n_2}{n_1}\\ {\theta }_{\text{c}}={\sin }^{-1}\left(\frac{n_2}{n_1}\right)\end{array}$                                                                                                        (I)

Here, ${\theta }_{\text{c}}$ is the critical angle and $n_1$ and $n_2$ are the refractive index of respective medium.

Write the expression for the index of refraction on the first face of the prism.

    $\frac{\sin i}{\sin r}=n$

Here, $i$ is the angle of incidence, $n$ is the index of refraction and $r$ is the angle of refraction.

Rewrite the above equation for $r$.

    $\begin{array}{c}\frac{\sin i}{\sin r}=n\\ \sin r=\frac{\sin i}{n}\\ r={\sin }^{-1}\left(\frac{\sin i}{n}\right)\end{array}$                                                                                                  (II)

Consider figure (1).

The angle of incidence is,

    $i=90°-r$                                                                                                                 (III)

Calculate the length $AB$.

    $\begin{array}{c}\tan (i)=\frac{AB}{AC}\\ AB=\left(\tan (i)\right)\left(AC\right)\end{array}$                                                                                              (IV)

The laser beam is hitting at the middle of the slab. Therefore the length $AC$ is half of the thickness of the slab. That is $AC=\frac{3.10\text{mm}}{2}$.

Half reflection is occurred at length $AB$, then total number of reflection is,

    $\begin{array}{c}\frac{\left(\frac{1}{2}\right)}{x}=\frac{AB}{AD}\\ x=\frac{AD}{2AB}\end{array}$                                                                                                                (V)

Here, $x$ is the total number of reflection.

Conclusion:

Substitute $1.48$ for $n_1$ and $1$ for $n_2$ in equation (I) to calculate ${\theta }_{\text{c}}$.

    $\begin{array}{c}{\theta }_{\text{c}}={\sin }^{-1}\left(\frac{1}{1.48}\right)\\ ={\sin }^{-1}\left(0.675\right)\\ =42.51°\end{array}$

Substitute $1.48$ for $n$ and $50.0°$ for $i$ in equation (II) to calculate $r$.

    $\begin{array}{c}r={\sin }^{-1}\left(\frac{\sin 50.0°}{1.48}\right)\\ ={\sin }^{-1}\left(\frac{0.766}{1.48}\right)\\ ={\sin }^{-1}\left(0.517\right)\\ =31.17°\end{array}$

Substitute $31.17°$ for $r$ in equation (III) to calculate $i$.

    $\begin{array}{c}i=90°-31.17°\\ =58.83°\end{array}$

Angle $i$ is greater than the angle ${\theta }_{\text{c}}$. Therefore, the ray will not escape the slab, and undergoes internal reflection.

Substitute $58.83°$ for $i$ and $\frac{3.10\text{mm}}{2}$ for $AC$ in equation (IV) to calculate $AB$.

    $\begin{array}{c}AB=\left(\tan (58.83°)\right)\left(\frac{3.10\text{mm}}{2}\right)\left(\frac{1\times {10}^{-3}\text{m}}{1\text{mm}}\right)\\ =\left(1.65\right)\left(\frac{3.10\times {10}^{-3}\text{m}}{2}\right)\\ =\left(1.65\right)\left(1.55\times {10}^{-3}\text{m}\right)\\ =2.56\times {10}^{-3}\text{m}\end{array}$

Substitute $2.56\times {10}^{-3}\text{m}$ for $AB$ and $42.0\text{cm}$ for $AD$ in equation (V) to calculate $x$.

    $\begin{array}{c}x=\frac{\left(42.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)}{2\left(2.56\times {10}^{-3}\text{m}\right)}\\ =\frac{\left(42.0\times {10}^{-2}\text{m}\right)}{\left(5.12\times {10}^{-3}\text{m}\right)}\\ =82\end{array}$

The total number of reflection possible is $82$, but it is given that $85$ reflection has occurred. Therefore given situation is impossible.