Chapter 38, Problem 125PQ

(a)

To determine

The proof for the statement, “the minimum deviation satisfies the equation $n=\frac{\sin \left(\frac{{\alpha }_{\text{min}}+\beta }{2}\right)}{\sin \left(\frac{\beta }{2}\right)}$.”

Answer to Problem 125PQ

The proof for the statement, “the minimum deviation satisfies the equation $n=\frac{\sin \left(\frac{{\alpha }_{\text{min}}+\beta }{2}\right)}{\sin \left(\frac{\beta }{2}\right)}$” is as stated below.

Explanation of Solution

Write the expression for refraction law at left side of the given prism.

    $\sin {\theta }_1=n\sin {\theta }_2$                                                                                                     (I)

Here, ${\theta }_1$ is incident angle of ray coming from left at left surface, ${\theta }_2$ is the refracted angle at left face of the prism and $n$ is refractive index of prism.

Write the expression for angle of deviation of the given prism.

    $\alpha ={\theta }_1+{\theta }_4-\beta$                                                                                                    (II)

Here, $\alpha$ is the angle of deviation and $\beta$ is the apex angle of the prism.

Write the relation among various angles from the geometry of the prism.

    ${\theta }_2+{\theta }_3=\beta$                                                                                                         (III)

Here, ${\theta }_3$ is incident angle of ray coming from left at right surface, ${\theta }_4$ is the refracted angle at left face of the prism

For minimum deviation, refracted angle at both faces must be equals.

    ${\theta }_2={\theta }_3$

For minimum deviation, incidence angle at both faces must be equal.

    ${\theta }_1={\theta }_4$

Substitute ${\theta }_2$ for ${\theta }_3$ in equation (III) to find ${\theta }_2$.

    $\begin{array}{l}{\theta }_2+{\theta }_2=\beta \\ {\theta }_2=\frac{\beta }{2}\end{array}$

Substitute ${\theta }_1$ for ${\theta }_4$ , ${\alpha }_{\text{min}}$ for $\alpha$ in equation (II) to find ${\theta }_1$.

    $\begin{array}{c}{\alpha }_{\text{min}}={\theta }_1+{\theta }_1-\beta \\ 2{\theta }_1=\alpha +\beta \\ {\theta }_1=\frac{{\alpha }_{\text{min}}+\beta }{2}\end{array}$

Here, ${\alpha }_{\text{min}}$ is the minimum deviation angle of the prism.

Further, substitute $\frac{{\alpha }_{\text{min}}+\beta }{2}$ for ${\theta }_1$ and $\frac{\beta }{2}$ for ${\theta }_2$ in equation (I) to find the refractive index.

    $\sin \left(\frac{{\alpha }_{\text{min}}+\beta }{2}\right)=n\sin \left(\frac{\beta }{2}\right)$

Therefore, the refractive index is,

    $n=\frac{\sin \left(\frac{{\alpha }_{\text{min}}+\beta }{2}\right)}{\sin \left(\frac{\beta }{2}\right)}$                                                                                               (V)

Conclusion:

Therefore, refractive index is $\frac{\sin \left(\frac{{\alpha }_{\text{min}}+\beta }{2}\right)}{\sin \left(\frac{\beta }{2}\right)}$.

(b)

To determine

The minimum deviation angle.

Answer to Problem 125PQ

The minimum deviation angle is ${\alpha }_{\text{min}}=\beta \left(n-1\right)$.

Explanation of Solution

Use the approximation method for angle.

    $\sin \beta \approx \beta$

Conclusion:

Substitute $\frac{{\alpha }_{\text{min}}+\beta }{2}$ for $\sin \left(\frac{{\alpha }_{\text{min}}+\beta }{2}\right)$ and $\frac{\beta }{2}$ for $\sin \left(\frac{\beta }{2}\right)$ in equation (V) to find the minimum deviation.

    $\begin{array}{c}n=\frac{\left(\frac{{\alpha }_{\text{min}}+\beta }{2}\right)}{\frac{\beta }{2}}\\ n\beta ={\alpha }_{\text{min}}+\beta \\ {\alpha }_{\text{min}}=\beta \left(n-1\right)\end{array}$

Therefore, minimum deviation angle is ${\alpha }_{\text{min}}=\beta \left(n-1\right)$.