Question

Shown from above in FIGURE P34.54 is one corner of a rectangular box filled with water. A laser beam starts $10\mathrm{cm}$from side A of the container and enters the water at position x. You can ignore the thin walls of the container.

a. If $\mathrm{x}=15\mathrm{cm}$, does the laser beam refract back into the air through side B or reflect from side B back into the water? Determine the angle of refraction or reflection.

b. Repeat part a for $\mathrm{x}=25\mathrm{cm}$.

c. Find the minimum value of x for which the laser beam passes through side B and emerges into the air.

Short answer

a. The laser beam will reflected back into the same water with angle $51.3^0.$

b. The laser will refracted into the air with the angle $72.2^0.$

c. The minimum value of x is$18\mathrm{cm}.$

Step-by-step solution

Part (a) Step 1: Given information

We have given,

A laser beam will enter from $10\mathrm{cm}$from the A into the water.

$\mathrm{x}=15\mathrm{cm}$

We have to find the angle for which it will either reflected or refracted from the point B.

Simplify

According to the Snell's law we can say,

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

Where,

${\mathrm{n}}_1$is the refractive index of air which is 1.

localid="1650639638343" ${\mathrm{n}}_2=1.33$the refractive index of water.

Let us consider that the laser fall on the glass such that it will make localid="1650638731877" $\mathrm{\phi }$angle with the side of the glass as shown.

Then, using the geometry we can write,

localid="1650638574608" $$\begin{gathered} \tan \phi =\frac{10cm}{x}=\frac{10}{15} \\ \phi =33.{69}^0 \end{gathered}$$

Then, the angle of incidence will be,

localid="1650638632058" $$\begin{gathered} {\mathrm{\theta }}_1={90}^0-33.{69}^0 \\ {\mathrm{\theta }}_1=56.3^0 \end{gathered}$$

Using the sell's law at the interface, we can write

localid="1650638690169" $$\begin{gathered} \sin {\theta }_2=\frac{n_1}{n_2}\sin {\theta }_1 \\ {\theta }_2={\sin }^{-1}\left(\frac{1}{1.33}\sin 56.3^0\right) \\ {\theta }_2=38.7^0 \end{gathered}$$

Now again using the geometry we can find,

localid="1650638754220" ${\mathrm{\theta }}_3={90}^0-38.7^0=51.3^0$

Then the critical angle will will found out a the interface,

localid="1650638803421" $$\begin{gathered} {\mathrm{\theta }}_{\mathrm{c}}={\sin }^{-1}\left(\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}\right) \\ {\mathrm{\theta }}_{\mathrm{c}}={\sin }^{-1}\left(\frac{1}{1.33}\right) \\ {\mathrm{\theta }}_{\mathrm{c}}=48.{75}^0 \end{gathered}$$

Since the angle is smaller then the laser beam will reflected back into the water with angle ${\mathrm{\theta }}_3.$from the law of reflection.

Part (b) Step 1: Given information

We have given,

The laser will fall at the interface at $10\mathrm{cm}$from the side.

$\mathrm{x}=25\mathrm{cm}$

We have to find the angle of refraction from the interface B.

Simplify

Similarly as part a we can use the geometry to find the ,

$$\begin{gathered} \tan \phi =\frac{10cm}{x}=\frac{10}{25} \\ \phi =21.8^0 \end{gathered}$$

Then, the angle of incidence will be,

$$\begin{gathered} {\mathrm{\theta }}_1={90}^0-21.8^0 \\ {\mathrm{\theta }}_1=68.2^0 \end{gathered}$$

Using the sell's law at the interface, we can write

$$\begin{gathered} \sin {\theta }_2=\frac{n_1}{n_2}\sin {\theta }_1 \\ {\theta }_2={\sin }^{-1}\left(\frac{1}{1.33}\sin 68.2^0\right) \\ {\theta }_2=44.{27}^0 \end{gathered}$$

Now again using the geometry we can find,

${\mathrm{\theta }}_3={90}^0-44.{27}^0=45.{72}^0$

Then the critical angle will will found out a the interface,

$$\begin{gathered} {\mathrm{\theta }}_{\mathrm{c}}={\sin }^{-1}\left(\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}\right) \\ {\mathrm{\theta }}_{\mathrm{c}}={\sin }^{-1}\left(\frac{1}{1.33}\right) \\ {\mathrm{\theta }}_{\mathrm{c}}=48.{75}^0 \end{gathered}$$

Since it is smaller than critical angle then it will refract into air.

then,

$$\begin{gathered} {\mathrm{\theta }}_4={\sin }^{-1}\left(\frac{1.33}{1}\sin 45.{72}^0\right) \\ {\mathrm{\theta }}_4=72.2^0 \end{gathered}$$

Part (c) Step 1: Given information

We have give o find the minimum value of x.

Simplify

To minimize the value of x let us consider that the ray will reflect through the interface as shown from side B.

Then, we will find the ${\mathrm{\theta }}_2$from the geometry is ,

$$\begin{gathered} {\mathrm{\theta }}_2={90}^0-{\mathrm{\theta }}_{\mathrm{c}} \\ {\mathrm{\theta }}_2={90}^0-48.{75}^0=41.{25}^0 \end{gathered}$$

Now using the Snell's law at the interface A is given by,

$$\begin{gathered} {\mathrm{\theta }}_1={\sin }^{-1}\left(\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\sin {\mathrm{\theta }}_2\right) \\ {\mathrm{\theta }}_1={\sin }^{-1}\left(\frac{1.33}{1}\sin 41.{25}^0\right) \\ {\mathrm{\theta }}_1=61.{27}^0 \end{gathered}$$

Then angle $\mathrm{\phi }$will becomes

$\mathrm{\phi }={90}^0-61.{27}^0=28.{73}^0$

Then x will found out as

$$\begin{gathered} \mathrm{x}=\frac{10\mathrm{cm}}{\mathrm{tan\phi }} \\ \mathrm{x}=\frac{10\mathrm{cm}}{\tan 28.{73}^0} \\ \mathrm{x}=18\mathrm{cm} \end{gathered}$$