Question

The refractive index of prism is µ = √3 and the ratio of the angle of minimum deviation to the angle of prism is one. The value of angle of prism is ______^o.

Answer 1

Correct Answer: 60

For $\delta_\text{min}$:

\[ i = e \quad \text{and} \quad r_1 = r_2 = \frac{A}{2} \]

\[ \delta_\text{min} = 2i - A \]

Given, $\frac{\delta_\text{min}}{A} = 1$:

\[ \frac{2i - A}{A} = 1 \]

Simplifying:

\[ 2i - A = A \implies 2i = 2A \implies i = A \]

Using Snell's law:

\[ 1 \cdot \sin i = \mu \cdot \sin r \implies \sin i = \mu \cdot \sin \left(\frac{A}{2}\right) \]

Substituting $i = A$:

\[ \sin A = \mu \cdot \sin \left(\frac{A}{2}\right) \]

Expanding $\sin A$:

\[ 2 \sin \frac{A}{2} \cos \frac{A}{2} = \sqrt{3} \cdot \sin \frac{A}{2} \]

Dividing by $\sin \frac{A}{2}$:

\[ 2 \cos \frac{A}{2} = \sqrt{3} \implies \cos \frac{A}{2} = \frac{\sqrt{3}}{2} \]

\[ \frac{A}{2} = 30^\circ \implies A = 60^\circ \]

Answer 2

This problem deals with the refraction of light through a prism. We are given the refractive index of the prism and the relationship between the angle of minimum deviation and the angle of the prism. The goal is to find the angle of the prism.

Concept Used:

The relationship between the refractive index (\(\mu\)), the angle of the prism (\(A\)), and the angle of minimum deviation (\(\delta_m\)) is given by the prism formula. This formula is derived from applying Snell's law at the two refracting surfaces of the prism under the condition of minimum deviation, which occurs when the angle of incidence equals the angle of emergence.

The prism formula is:

\[ \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

Step-by-Step Solution:

Step 1: List the given information from the problem statement.

The refractive index of the prism, \(\mu = \sqrt{3}\).

The ratio of the angle of minimum deviation to the angle of the prism is one. This can be written as:

\[ \frac{\delta_m}{A} = 1 \implies \delta_m = A \]

Step 2: Substitute the given relationship \(\delta_m = A\) into the prism formula.

\[ \mu = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

Simplifying the numerator:

\[ \mu = \frac{\sin\left(\frac{2A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin(A)}{\sin\left(\frac{A}{2}\right)} \]

Step 3: Use the trigonometric double-angle identity for sine, which is \(\sin(A) = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\).

Substituting this identity into the equation:

\[ \mu = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

Step 4: Cancel the term \(\sin\left(\frac{A}{2}\right)\) from the numerator and the denominator.

\[ \mu = 2\cos\left(\frac{A}{2}\right) \]

Step 5: Substitute the given value of the refractive index \(\mu = \sqrt{3}\) into the simplified equation and solve for \(A\).

\[ \sqrt{3} = 2\cos\left(\frac{A}{2}\right) \] \[ \cos\left(\frac{A}{2}\right) = \frac{\sqrt{3}}{2} \]

Step 6: Determine the angle \(\frac{A}{2}\) from the value of its cosine.

We know that \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\). Therefore:

\[ \frac{A}{2} = 30^\circ \]

Final Computation & Result:

To find the angle of the prism, \(A\), we multiply the result from Step 6 by 2.

\[ A = 2 \times 30^\circ \] \[ A = 60^\circ \]

The value of the angle of prism is 60°.