Question

Two parallel rays are incident on two surfaces of a prism of prism angle $\theta$ = 45° and get reflected. Angle of diversion between two rays is $\varphi$. If angle of incidence of both ray are equal and is $\alpha$.

• A. 90
• B. 45
• C. 135
• D. 180

Answer 1

Answer: (A)

90

Answer 2

Let the incident rays be parallel. Let the angle of incidence on both faces be $\alpha$. Let the normals to the two faces be $\vec{n_1}$ and $\vec{n_2}$. The angle between the normals is $180^\circ - \theta$. Since the incident rays are parallel and the angle of incidence is the same for both, the direction of the incident rays bisects the angle between the normals. Let the incident ray direction be along the x-axis. Let the normal to the first face, $\vec{n_1}$, make an angle $\psi$ with the x-axis. Then, the angle of incidence $\alpha = \psi$. The reflected ray from the first face will make an angle $2\psi = 2\alpha$ with the x-axis.

Let the normal to the second face, $\vec{n_2}$, make an angle $-\psi$ with the x-axis. Then, the angle of incidence $\alpha = \psi$. The reflected ray from the second face will make an angle $-2\psi = -2\alpha$ with the x-axis.

The angle between the two normals is $2\psi$. We know that the angle between the normals is $180^\circ - \theta$. So, $2\psi = 180^\circ - \theta$. This implies $\psi = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{\theta}{2}$. Since $\alpha = \psi$, we have $\alpha = 90^\circ - \frac{\theta}{2}$.

The angle between the two reflected rays is $\varphi$. $\varphi = |(2\psi) - (-2\psi)| = |4\psi|$. Substituting $\psi = 90^\circ - \frac{\theta}{2}$: $\varphi = |4 \left( 90^\circ - \frac{\theta}{2} \right)| = |360^\circ - 2\theta|$.

Let's re-evaluate the relationship between the prism angle and the angle between normals. For a prism, the angle between the normals to the two refracting surfaces is equal to the prism angle $\theta$. So, the angle between $\vec{n_1}$ and $\vec{n_2}$ is $\theta$. Since the incident ray direction bisects this angle, the angle between the incident ray and each normal is $\alpha = \theta/2$.

Now, consider the reflected rays. Let the incident ray be along the x-axis. Let the normal $\vec{n_1}$ be at an angle $\alpha$ with the x-axis. The reflected ray makes an angle $2\alpha$ with the x-axis.

Let the normal $\vec{n_2}$ be at an angle $-\alpha$ with the x-axis. The reflected ray makes an angle $-2\alpha$ with the x-axis.

The angle between the two reflected rays is $\varphi = |2\alpha - (-2\alpha)| = |4\alpha|$. Since $\alpha = \theta/2$: $\varphi = |4 (\theta/2)| = |2\theta|$.

Given $\theta = 45^\circ$: $\varphi = 2 \times 45^\circ = 90^\circ$.