The refractive index of a prism with apex angle A is ( A2). The angle of minimum deviation is:

Given: Refractive index of the prism is n = ( A2). Find: The angle of minimum deviation m. For a prism at minimum deviation, n = ( A + m2) ( A2) Using the given value, ( A2) = ( A2) ( A2) So, ( A + m2) ( A2) = ( A2) ( A2) Therefore, ( A + m2) = ( A2) Using = (90^ -), we get A + m2 = 90^ - A2 Now solve: A + m = 180^ - A m = 180^ - 2A Therefore, the correct option is D. Given: n = ( A2). Find: m. Start from n = ( A + m2) ( A2) Substitute n = ( A2): ( A + m2) ( A2) = ( A2) ( A2) Hence, ( A + m2) = ( A2) Since x = (90^ - x), compare the angles directly: A + m2 = 90^ - A2 So, m = 180^ - 2A Thus, the angle of minimum deviation is 180^ - 2A.

The refractive index of a prism with apex angle A is ( A2). The a… — Solution | JEE Physics

MCQEasyJEE 2024Prisms & Total Internal Reflection

The refractive index of a prism with apex angle $A$ is $\cot\left(\frac{A}{2}\right)$ . The angle of minimum deviation is:

A
$\delta_m = 180^\circ - A$
B
$\delta_m = 180^\circ - 3A$
C
$\delta_m = 180^\circ - 4A$
D
$\delta_m = 180^\circ - 2A$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Refractive index of the prism is $n = \cot\left(\frac{A}{2}\right)$ .

Find: The angle of minimum deviation $\delta_m$ .

For a prism at minimum deviation,

n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

Using the given value,

\cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}

So,

\frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}

Therefore,

\sin\left(\frac{A + \delta_m}{2}\right) = \cos\left(\frac{A}{2}\right)

Using $\sin\theta = \cos(90^\circ - \theta)$ , we get

\frac{A + \delta_m}{2} = 90^\circ - \frac{A}{2}

Now solve:

A + \delta_m = 180^\circ - A

\delta_m = 180^\circ - 2A

Therefore, the correct option is D.

Direct Identity Approach

Given: $n = \cot\left(\frac{A}{2}\right)$ .

Find: $\delta_m$ .

Start from

n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

Substitute $n = \cot\left(\frac{A}{2}\right)$ :

\frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}

Hence,

\sin\left(\frac{A + \delta_m}{2}\right) = \cos\left(\frac{A}{2}\right)

Since $\cos x = \sin(90^\circ - x)$ , compare the angles directly:

\frac{A + \delta_m}{2} = 90^\circ - \frac{A}{2}

So,

\delta_m = 180^\circ - 2A

Thus, the angle of minimum deviation is $180^\circ - 2A$ .

Common mistakes

Using the prism formula incorrectly by writing $n = \frac{\sin A + \delta_m}{2\sin(A/2)}$ or other misplaced forms. This is wrong because the sine applies to the entire angle $\frac{A + \delta_m}{2}$ . Always use the standard relation exactly as $n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ .
Not converting $\cot\left(\frac{A}{2}\right)$ into $\frac{\cos(A/2)}{\sin(A/2)}$ . This is wrong because the cancellation with the denominator $\sin(A/2)$ becomes unclear. Rewrite cotangent in sine-cosine form before equating.
Using the identity $\sin\theta = \cos(90^\circ - \theta)$ incorrectly and taking $\frac{A + \delta_m}{2} = \frac{A}{2}$ . This is wrong because $\cos\left(\frac{A}{2}\right) = \sin\left(90^\circ - \frac{A}{2}\right)$ , not $\sin\left(\frac{A}{2}\right)$ . Match the complementary angle carefully.

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