Question

Question: In the Figure, light enters a ${\mathbf{90}}^{\mathbf{\circ }}$triangular prism at point Pwith incident angle $\mathit{\theta }$, and then some of it refracts at point Qwith an angle of refraction of ${\mathbf{90}}^{\mathbf{\circ }}$.(a) What is the index of refraction of the prism in terms of $\mathit{\theta }$?(b) What, numerically, is the maximum value that the index of refraction can have?(c) Does light emerge at Qif the incident angle at Pis increased slightly? and (d) Does light emerge at Qif the incident angle at Pis decreased slightly?

Figure:

Short answer

1. Refractive index of the prism in terms of $\theta$islocalid="1662993422145" $n=\sqrt{1+{\sin }^2\theta }$.
2. Value for the maximum refractive index is $1.41$.
3. Yes; light will emerge at Q if the incident angle P is increased slightly.
4. No; light does not emerge at Q if the incident angle P is decreased slightly.

Step-by-step solution

Given

1. Angle of incidence at point P =$\theta$
2. The angle of refraction at Q is${\theta }_4={90}^{\circ }$

Understanding the concept

Draw a ray diagram for the given case. From Snell’s law, we can find the formula for the refractive index of prism in terms of $\theta$. From this formula, we can get the condition for the maximum refractive index. Using this, obtain the value for the maximum refractive index. Whether light emerges depends on whether the angle of incidence at point Q is greater than or less than the critical angle.

Formula

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

(a) Calculate the refractive of index of the prism in terms of θ

The ray diagram for the given case is as follows:

$\mathrm{Let}{\mathrm{n}}_2=n_3=n$and $\theta ={\theta }_1$

Applying Snell’s law at point P, we get

$n_{1}sin{\theta }_1=n_2\sin {\theta }_2$

But,

$n_1=1$

Hence,

$$\begin{gathered} {\mathrm{sin\theta }}_1={\mathrm{n}}_2{\mathrm{sin\theta }}_2 \\ {\mathrm{sin\theta }}_2=\frac{{\mathrm{sin\theta }}_1}{{\mathrm{n}}_2} \\ {\mathrm{sin\theta }}_2=\frac{{\mathrm{sin\theta }}_1}{\mathrm{n}} \end{gathered}$$

From the diagram, write the equation for the angle as:

$$\begin{gathered} {\theta }_2+{\theta }_3={90}^{\circ } \\ {\theta }_3={90}^{\circ }-{\theta }_2 \end{gathered}$$

Apply Snell’s law at point Q, and solve as:

${\mathrm{n}}_3{\mathrm{sin\theta }}_3={\mathrm{n}}_4{\mathrm{sin\theta }}_4$

But,

$n_4=1$

Hence,

$$\begin{gathered} {\mathrm{n}}_3{\mathrm{sin\theta }}_3=1 \\ {\mathrm{n}}_3\sin \left(90-{\mathrm{\theta }}_2\right)=1 \\ {\mathrm{n}}_3{\mathrm{cos\theta }}_2=1 \\ {\mathrm{n}}_3\sqrt{1-{\sin }^2{\mathrm{\theta }}_2}=1 \end{gathered}$$

Solve further as:

n$$\begin{gathered} n\sqrt{1-{\sin }^2{\theta }_2}=1 \\ n\sqrt{1-{\left(\frac{\sin {\theta }_2}{n}\right)}^2}=1 \\ n\times \frac{\sqrt{n^2-{\sin }^2{\theta }_1}}{n}=1 \\ \sqrt{n^2-{\sin }^2{\theta }_1}=1 \end{gathered}$$

Take square roots on both sides to get:

$$\begin{gathered} n^2-{\sin }^2{\theta }_1=1 \\ {n}^2=1+{\sin }^2{\theta }_1 \\ n=\sqrt{1+{\sin }^2{\theta }_1}=\sqrt{1+{\sin }^2\theta } \end{gathered}$$

(b) Calculate the value for the maximum refractive index

For maximum value of the refractive index,

Value for${\sin }^2\mathrm{\theta }=1$

Substitute the values in the above equation and solve as:

$n=\sqrt{1+1}=\sqrt{2}=1.41$

(c) Find out if the light emerges at Q if the incident angle P is increased slightly

Refractive index value is constant, so from part (A), we can conclude that an increase in the value for${\theta }_1$will cause an increase in the value for${\theta }_2$.

If${\theta }_2$increases, this will lead to a decrease in the value for${\theta }_3$and this is less than the critical angle.

From this, we can conclude that the ray will emerge out of the prism.

(d) Find out if the light emerges at Q if the incident angle P is decreased slightly

Refractive index value is constant, so a decrease in the value for ${\theta }_1$will cause a decrease in the value for ${\theta }_2$ .

If ${\theta }_2$decreases, this will lead to an increase in the value for ${\theta }_3$and this is greater than the critical angle. So, it undergoes total internal reflection.

From this, it is concluded that the ray will get reflected into the prism.