Physics Q42 – JEE-MAIN 2026

Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is _______.

Clue 1 of 3: Prism Properties

Recall the properties of an equilateral prism and the formula for refractive index at minimum deviation.

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Step 1: Identify Prism Parameters✦ Active

For an equilateral prism, the angle of the prism $A$ is $60^{\circ}$ . The problem states that the angle of minimum deviation $δ_{m}$ is half of the angle of the prism.

A = 60^{\circ}

δ_{m} = \frac{A}{2} = \frac{60^{\circ}}{2} = 30^{\circ}

Step 2: Apply Refractive Index Formula○ Expand

The formula for the refractive index $μ$ of a prism at minimum deviation is:

μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}

💡 Teacher's Secret Hint

Ensure correct substitution of $A$ and $δ_{m}$ into the formula.

Step 3: Calculate Refractive Index○ Expand

Substitute the values of $A$ and $δ_{m}$ into the formula and simplify:

μ = \frac{\sin (\frac{60^{\circ} + 30^{\circ}}{2})}{\sin (\frac{60^{\circ}}{2})} = \frac{\sin (\frac{90^{\circ}}{2})}{\sin (30^{\circ})} = \frac{\sin (45^{\circ})}{\sin (30^{\circ})}

μ = \frac{1 / \sqrt{2}}{1 / 2} = \frac{1}{\sqrt{2}} \times 2 = \frac{2}{\sqrt{2}} = \sqrt{2}

Comparing this result with the given options, $\sqrt{2}$ corresponds to option 3.

💡 Teacher's Secret Hint

Remember the standard trigonometric values for $30^{\circ}$ and $45^{\circ}$ .

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Physics Question 42 – JEE-MAIN 2026

Clue 1 of 3: Prism Properties

🧠 Full Solution Path