If the refractive index of the material of a prism is ( A2), where A is the angle of the prism, then the angle of minimum deviation will be:

If the refractive index of the material of a prism is ( A2), wher… — Solution | JEE Physics

MCQEasyJEE 2024Prisms & Total Internal Reflection

If the refractive index of the material of a prism is $\cot\left(\frac{A}{2}\right)$ , where $A$ is the angle of the prism, then the angle of minimum deviation will be:

A
$\pi - 2A$
B
$\frac{\pi}{2} - 2A$
C
$\pi - A$
D
$\frac{\pi}{2} - A$

Answer

Correct answer:B

Step-by-step solution

Standard Method

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

Find: The angle of minimum deviation.

For a prism at minimum deviation,

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

Substituting $n = \cot\left(\frac{A}{2}\right)$ ,

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

Using

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

we get

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

Cancelling $\sin\left(\frac{A}{2}\right)$ from both sides,

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

Now use the identity

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

Hence,

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

Equating the angles,

\frac{A + \delta_m}{2} = \frac{\pi}{2} - \frac{A}{2}

So,

A + \delta_m = \pi - A

Therefore,

\delta_m = \pi - 2A

Therefore, the angle of minimum deviation is $\pi - 2A$ . The solution working gives this result, although the solution incorrectly marks the correct option as B. Hence the defensible option is A.

Detailed Algebra

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

Find: $\delta_{\min}$

Start with the prism formula at minimum deviation:

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

Insert the given value of refractive index:

\cot\left(\frac{A}{2}\right) = \frac{\sin\left(\frac{A + \delta_{\min}}{2}\right)}{\sin\left(\frac{A}{2}\right)}

Rewrite cotangent:

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

After cancellation,

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

Convert cosine into sine form:

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

Thus,

\sin\left(\frac{A + \delta_{\min}}{2}\right) = \sin\left(\frac{\pi}{2} - \frac{A}{2}\right)

Taking the principal equality used in the provided solution,

\frac{A + \delta_{\min}}{2} = \frac{\pi}{2} - \frac{A}{2}

which gives

\delta_{\min} = \pi - 2A

So the correct option from the listed choices is A.

Common mistakes

Using the prism formula incorrectly by writing $n = \frac{\sin A + \delta_m}{2 \sin(A/2)}$ . This is wrong because the sine must act on the entire angle $\frac{A + \delta_m}{2}$ . Use the correct formula with proper brackets.
Not converting $\cot\left(\frac{A}{2}\right)$ into $\frac{\cos(A/2)}{\sin(A/2)}$ before cancellation. Without this step, the simplification cannot be justified. Rewrite trigonometric ratios explicitly first.
Choosing option B only from the solution. This is wrong because the actual solution working ends with $\delta_m = \pi - 2A$ , which matches option A. Always trust the derivation over a contradictory label.

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