Question: In Fig. 33-62, a light ray in air is incident at angle ${\mathbf{\theta }}_{\mathbf{1}}$on a block of transparent plastic with an index of refraction of. The dimensions indicated are $\mathbf{H}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$and$\mathbf{W}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$. The light passes through the block to one of its sides and there undergoes reflection (inside the block) and possibly refraction (out into the air). This is the point offirst reflection. The reflected light then passes through the block to another of its sides — a point ofsecond reflection. If ${\mathbf{\theta }}_{\mathbf{1}}\mathbf{=}{\mathbf{40}}^{\mathbf{\circ }}$, on which side is the point of (a) first reflection and (b) second reflection? If there is refraction at the point of (c) first reflection, and (d) second reflection, give the angle of refraction; if not, answer “none”. If ${\mathbf{\theta }}_{\mathbf{1}}\mathbf{=}{\mathbf{70}}^{\mathbf{\circ }}$, on which side is the point of (e) first reflection and (f) second reflection? If there is refraction at the point of (g) first reflection, and (h) second reflection, give the angle of refraction; if not, answer “none”.

Dimensions of the block are, $\mathrm{H}=2.00\mathrm{cm}$and$\mathrm{W}=3.00\mathrm{cm}$

$$\begin{gathered} {\mathrm{\theta }}_1={40}^{\circ } \\ {\mathrm{\theta }}_2={70}^{\circ } \end{gathered}$$

The Refractive Index of plastic is $1.56$.

We can find the angle of incidence and reflection by using Snell’s law. Then by using trigonometry, we can calculate the distance traveled by the ray and then compare it with the dimensions of the block to determine on which side the ray strikes.

Formula:

Snell’s law:

${\mathrm{n}}_1{\mathrm{sin\theta }}_1={\mathrm{n}}_2{\mathrm{sin\theta }}_2$

To find the first point of reflection:

Applying Snell’s law to determine the angle of refraction:

$$\begin{gathered} n_1\sin {\theta }_1=n_2\sin {\theta }_2 \\ 1.0\times \sin {40}^{\circ }=1.56\times \sin {\theta }_2 \\ {\theta }_2={\sin }^{-1}\left(\frac{\sin {40}^{\circ }}{1.56}\right) \\ {\theta }_2=24.{34}^{\circ } \end{gathered}$$

Now we can find the vertical distance traveled by the ray.

$\begin{array}{l}tan\left({\theta }_2\right)=\frac{y}{W}\\ y=3.00\mathrm{cm}\times tan(24.{34}^{\circ })\\ y=1.36\mathrm{cm}\end{array}$

This is less than H, so refracted ray strikes the side 3.

To find the second reflection point:

We know that for reflection, the angle is symmetrical.

Now find the horizontal distance traveled by the reflected ray.

$$\begin{gathered} {\mathrm{tan\theta }}_2=\frac{\mathrm{H}-1.36}{\mathrm{x}} \\ \mathrm{x}=\frac{2.00\mathrm{cm}-1.36\mathrm{cm}}{\tan 24.{34}^{\circ }} \\ \mathrm{x}=1.42\mathrm{cm} \end{gathered}$$

So, this ray travels$1.42\mathrm{cm}$to the left of the right corner, which is less than W.

Hence, it strikes the side 2.

To check whether there is refraction, first, we have to find the critical angle for the total internal reflection.

$\begin{array}{rcl}{\theta }_c& =& {\sin }^{-1}\left(\frac{n_{air}}{n_r}\right)\\ & =& {\sin }^{-1}\left(\frac{1.00}{1.56}\right)\\ & =& 39.9^{\circ }\end{array}$

Since the angle of incidence for refraction at side 3 is less than this.

Hence, refraction will occur at the point

To find the angle of incidence for refraction at the point

Here normal axis is vertical. By using trigonometry:

$\begin{array}{rcl}\mathrm{Angle}\mathrm{of}\mathrm{incidence}& =& {90}^{\circ }-{\mathrm{\theta }}_2\\ & =& {90}^{\circ }-24.{34}^{\circ }\\ & =& 65.{66}^{\circ }\\ & & \end{array}$

To find the first point of reflection:

Applying Snell’s law to determine the angle of refraction:

$$\begin{gathered} n_1\sin {\theta }_1=n_2\sin {\theta }_2 \\ 1.0\times \sin {70}^{\circ }=1.56\times \sin {\theta }_2 \\ {\theta }_2={\sin }^{-1}\left(\frac{\sin {70}^{\circ }}{1.56}\right) \\ {\theta }_2=37.{05}^{\circ } \end{gathered}$$

Now we can find the horizontal distance traveled by the ray.

$\begin{array}{l}tan\left({\mathrm{\theta }}_2\right)=\frac{\mathrm{H}}{\mathrm{x}}\\ \mathrm{x}=\frac{2.00\mathrm{cm}}{\tan \left(37.{05}^{\circ }\right)}\\ \mathrm{x}=2.65\mathrm{cm}\end{array}$

This is less than W so refracted ray strikes the side 2.

To find the second reflection point:

Here, the normal axis is vertical.

$\begin{array}{rcl}Angleofincidence& =& {90}^{\circ }-{\theta }_2\\ & =& {90}^{\circ }-37.{05}^{\circ }\\ & =& 52.{95}^{\circ }\end{array}$

Now find the vertical distance traveled by the reflected ray.

$$\begin{gathered} \mathrm{tan\theta }=\frac{\mathrm{W}-2.65}{\mathrm{y}} \\ \mathrm{y}=\frac{3.00\mathrm{cm}-2.65\mathrm{cm}}{\tan 52.{95}^{\circ }} \\ \mathrm{y}=0.264\mathrm{cm} \end{gathered}$$

So, this ray travel below from the upper right corner, which is less than .

Hence, it strikes the side .

To check whether there is refraction, first, we have to find the critical angle for the total internal reflection.

$\begin{array}{rcl}{\theta }_c& =& {\sin }^{-1}\left(\frac{n_{air}}{n_r}\right)\\ & =& {\sin }^{-1}\left(\frac{1}{1.56}\right)\\ & =& 39.9^{\circ }\end{array}$

And the angle of incidence for refraction at side 2 is greater than this. Hence there will be no refraction at point 1.

To find the angle of incidence for refraction at the point

Here normal axis is vertical. By using trigonometry:

$\begin{array}{rcl}\mathrm{Angle}\mathrm{of}\mathrm{incidence}& =& 90°-\mathrm{\theta }\\ & =& 90°-52.{95}^{\circ }\\ & =& 37.{05}^{\circ }\end{array}$

This is less than the critical angle, so refraction occurs at this point with the same angle${70}^{\circ }$.