Question

A ray is incident on an equilateral prism. The refractive index of the material of the prism is $\sqrt{\frac{7}{3}}$. Find the maximum deviation considering no total internal reflection in the prism

Answer 1

60 degrees

Answer 2

To find the maximum deviation, we need to consider the extreme possible angles of incidence for which the ray can emerge from the prism without undergoing total internal reflection (TIR).

1. Prism Angle (A): For an equilateral prism, $A = 60^\circ$.

2. Refractive Index ($\mu$): Given $\mu = \sqrt{\frac{7}{3}}$.

3. Critical Angle (C): The critical angle for the prism-air interface is given by $\sin C = \frac{1}{\mu}$. $\sin C = \frac{1}{\sqrt{7/3}} = \sqrt{\frac{3}{7}}$. $C = \arcsin\left(\sqrt{\frac{3}{7}}\right)$. Numerically, $\sqrt{\frac{3}{7}} \approx \sqrt{0.42857} \approx 0.6547$. Since $\sin 30^\circ = 0.5$ and $\sin 45^\circ = 0.707$, $C$ is between $30^\circ$ and $45^\circ$. $C \approx 40.89^\circ$.

4. Condition for No Total Internal Reflection: For the ray to emerge from the second face, the angle of incidence inside the prism at the second face ($r_2$) must be less than or equal to the critical angle $C$. So, $r_2 \le C$.

5. Prism Geometry and Snell's Law:

   • $A = r_1 + r_2$ (where $r_1$ and $r_2$ are angles of refraction inside the prism at the first and second faces, respectively).
   • $\sin i_1 = \mu \sin r_1$ (at the first face).
   • $\mu \sin r_2 = \sin i_2$ (at the second face).
   • Deviation $\delta = (i_1 + i_2) - A$.

6. Maximizing Deviation: The deviation $\delta$ is maximum when the angles of incidence ($i_1$) and emergence ($i_2$) are at their extreme possible values. These extremes are usually $0^\circ$ or $90^\circ$. The deviation function $\delta(i_1)$ has a minimum and increases as $i_1$ moves away from the minimum deviation condition. Thus, the maximum deviation occurs at the boundaries of the allowed range of $i_1$.

   The allowed range of $i_1$ is determined by two conditions:

   • $0^\circ \le i_1 \le 90^\circ$.
   • The ray must emerge from the second face (i.e., $r_2 \le C$).

   From $A = r_1 + r_2$, we have $r_2 = A - r_1$. So, the condition $r_2 \le C$ becomes $A - r_1 \le C$, which implies $r_1 \ge A - C$. Also, $r_1 \le C$ (since $i_1 \le 90^\circ \implies \sin r_1 = \frac{\sin i_1}{\mu} \le \frac{1}{\mu} = \sin C \implies r_1 \le C$). So, $r_1$ must be in the range $[A-C, C]$. This implies $i_1$ must be in the range $[\arcsin(\mu \sin(A-C)), \arcsin(\mu \sin C)]$. Let's evaluate the boundaries:

   • Upper boundary for $i_1$: $\mu \sin C = \mu \cdot \frac{1}{\mu} = 1$. So, $\arcsin(1) = 90^\circ$. This means $i_1 = 90^\circ$ is a possible angle of incidence.
   • Lower boundary for $i_1$ (for emergence): We need to calculate $\mu \sin(A-C)$. $\sin(A-C) = \sin(60^\circ - C) = \sin 60^\circ \cos C - \cos 60^\circ \sin C$. We know $\sin C = \frac{1}{\mu}$ and $\cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - \frac{1}{\mu^2}} = \frac{\sqrt{\mu^2-1}}{\mu}$. $\sin(A-C) = \frac{\sqrt{3}}{2} \frac{\sqrt{\mu^2-1}}{\mu} - \frac{1}{2} \frac{1}{\mu} = \frac{1}{2\mu} (\sqrt{3}\sqrt{\mu^2-1} - 1)$. Substitute $\mu = \sqrt{7/3}$: $\mu^2 = 7/3 \implies \mu^2 - 1 = 7/3 - 1 = 4/3$. $\sqrt{\mu^2-1} = \sqrt{4/3} = \frac{2}{\sqrt{3}}$. $\sin(A-C) = \frac{1}{2\sqrt{7/3}} \left(\sqrt{3} \cdot \frac{2}{\sqrt{3}} - 1\right) = \frac{\sqrt{3}}{2\sqrt{7}} (2 - 1) = \frac{\sqrt{3}}{2\sqrt{7}}$. Now, calculate $\mu \sin(A-C) = \sqrt{\frac{7}{3}} \cdot \frac{\sqrt{3}}{2\sqrt{7}} = \frac{1}{2}$. So, the minimum angle of incidence for emergence is $\arcsin(1/2) = 30^\circ$.

   Thus, the ray can emerge for $i_1$ in the range $[30^\circ, 90^\circ]$. The maximum deviation will occur at either $i_1 = 30^\circ$ or $i_1 = 90^\circ$.

7. Calculate Deviation for Extreme Cases:

   • Case 1: $i_1 = 90^\circ$ (grazing incidence) Using $\sin i_1 = \mu \sin r_1$: $\sin 90^\circ = \sqrt{7/3} \sin r_1 \implies 1 = \sqrt{7/3} \sin r_1 \implies \sin r_1 = \sqrt{3/7}$. So, $r_1 = C \approx 40.89^\circ$. Now find $r_2$: $r_2 = A - r_1 = 60^\circ - C = 60^\circ - 40.89^\circ = 19.11^\circ$. Since $r_2 = 19.11^\circ < C \approx 40.89^\circ$, there is no TIR. Now find $i_2$ using $\mu \sin r_2 = \sin i_2$: $\sin i_2 = \sqrt{7/3} \sin(60^\circ - C) = \sqrt{7/3} \cdot \frac{1}{2} = \frac{1}{2}$. So, $i_2 = 30^\circ$. Deviation $\delta = i_1 + i_2 - A = 90^\circ + 30^\circ - 60^\circ = 120^\circ - 60^\circ = 60^\circ$.

   • Case 2: $i_1 = 30^\circ$ (minimum incidence for emergence) This case is symmetric to Case 1. If $i_1 = 30^\circ$, then $r_1 = A-C$. This implies $r_2 = A - r_1 = A - (A-C) = C$. If $r_2 = C$, then $\sin i_2 = \mu \sin C = \mu \cdot \frac{1}{\mu} = 1$. So, $i_2 = 90^\circ$ (grazing emergence). Deviation $\delta = i_1 + i_2 - A = 30^\circ + 90^\circ - 60^\circ = 120^\circ - 60^\circ = 60^\circ$.

Both extreme cases yield the same maximum deviation.