Question

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index \(n\) of the prism satisfies:

• A. \(\sqrt{2}<n<2\)
• B. \(\sqrt{2}<n<2\sqrt{2}\)
• C. \(n\ge 2\)
• D. \(1<n<2\)

Hint:

When minimum deviation equals prism angle, always substitute \(\delta=A\) directly into the prism formula before simplifying.

Answer 1

The Correct Option is A

The problem describes the scenario of a transparent prism where the angle of minimum deviation is equal to the refracting angle of the prism. To solve this problem, let's explore the relevant concepts and derive the necessary formulae.

The refractive index \(n\) of a prism can be determined using the formula for the angle of minimum deviation:

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

where:

• \(A\) is the refracting angle of the prism.
• \(D_m\) is the angle of minimum deviation.

According to the problem, \(D_m = A\). Substituting this into the formula, we get:

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

Using the sine double-angle formula, \(\sin(A) = 2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right)\), the expression becomes:

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

The value of \(\cos\left(\frac{A}{2}\right)\) ranges from \(0\) to \(1\), implying:

\(1 < n \leq 2\)

However, given the options and the possibility of an ideal prism, we generally accept a value within:

\(\sqrt{2} < n < 2\)

Therefore, the correct answer is \(\sqrt{2} < n < 2\).

Let's evaluate the options to ensure this interpretation:

• \(\sqrt{2}\)
• \(\sqrt{2}\)
• \(n\ge 2\): Incorrect, since the maximum possible value is \(2\).
• \(1\)

Thus, the correct answer is confirmed: \(\sqrt{2}\)