Why can we write cos(A/2) as sin(90° - A/2)?

Because sin(90° - θ) = cosθ for any angle θ in the primary quadrant.

JEE MainPhysics

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Angle of Minimum Deviation for a Prism with Cotangent Refractive Index

The refracting angle of a prism is A and refractive index of the material of the prism is cot(A/2). Then the angle of minimum deviation will be -

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A180 - 2AB90 - AC180 + 2AD180 - 3A

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Question type: Single correct
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

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Students also ask

Why can we write

\cos (A / 2)

\sin (90^{\circ} - A / 2)

Because $\sin (90^{\circ} - θ) = \cos θ$ for any angle $θ$ in the primary quadrant.

Answer

Using the prism formula $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (A / 2)}$ with $μ = \cot (A / 2) = \frac{\cos (A / 2)}{\sin (A / 2)}$ , equating $\sin (\frac{A + δ_{m}}{2}) = \cos (A / 2) = \sin (90^{\circ} - A / 2)$ gives $δ_{m} = 180^{\circ} - 2 A$ .

Why each option works or fails

A · correct: 180 - 2A - This is the correct result derived by equating the prism formula to $\cot (A / 2)$ and converting cosine to sine. Correctly equating $\frac{A + δ_{m}}{2} = 90^{\circ} - \frac{A}{2}$ yields $δ_{m} = 180^{\circ} - 2 A$ .
B: 90 - A - The student forgets to multiply both sides of $\frac{A + δ_{m}}{2} = 90^{\circ} - \frac{A}{2}$ by $2$ , or mistakes $\frac{A + δ_{m}}{2}$ directly for $δ_{m}$ . Multiply the entire argument $90^{\circ} - \frac{A}{2}$ by $2$ to isolate $A + δ_{m}$ , giving $2$ .
C: 180 + 2A - The student makes a sign error when transposing the angle $A$ from the left-hand side to the right-hand side. Subtracting $A$ from both sides of $A + δ_{m} = 180^{\circ} - A$ gives $δ_{m} = 180^{\circ} - 2 A$ , not $180^{\circ} + 2 A$ .
D: 180 - 3A - The student doubles the term $A$ twice or adds an extra $- A$ during intermediate algebraic manipulation. Follow the linear step carefully: $A + δ_{m} = 180^{\circ} - A ⟹ δ_{m} = 180^{\circ} - A - A = 180^{\circ} - 2 A$ .

Step-by-step solution

Given: Prism refracting angle $= A$ , refractive index $μ = \cot (A / 2)$ .
Find: The angle of minimum deviation $δ_{m}$ .
Strategise: Use the prism formula $μ = \frac{\sin ((A + δ_{m}) / 2)}{\sin (A / 2)}$ . Substitute $μ = \cot (A / 2) = \frac{\cos (A / 2)}{\sin (A / 2)}$ and equate arguments using $\cos θ = \sin (90^{\circ} - θ)$ .
Execute: Equating expressions: $\frac{\sin ((A + δ_{m}) / 2)}{\sin (A / 2)} = \frac{\cos (A / 2)}{\sin (A / 2)} ⟹ \sin (\frac{A + δ_{m}}{2}) = \cos (\frac{A}{2}) = \sin (90^{\circ} - \frac{A}{2})$ . Thus, $\frac{A + δ_{m}}{2} = 90^{\circ} - \frac{A}{2} ⟹ A + δ_{m} = 180^{\circ} - A ⟹ δ_{m} = 180^{\circ} - 2 A$ .
Verify: For an equilateral prism $A = 60^{\circ}$ , $μ = \cot (30^{\circ}) = \sqrt{3} \approx 1.732$ . Formula gives $δ_{m} = 180^{\circ} - 2 (60^{\circ}) = 60^{\circ}$ . By standard prism formula: $μ = \sin (60^{\circ}) / \sin (30^{\circ}) = (\sqrt{3} / 2) / (1 / 2) = \sqrt{3}$ , which matches exactly.

Shortcut:

When to use it: Fastest verification method in MCQs with symbolic options.

Given: Choose a convenient angle $A = 60^{\circ}$ . Then $μ = \cot (30^{\circ}) = \sqrt{3}$ .

Strategise: Substitute $μ = \sqrt{3}$ and $A = 60^{\circ}$ into $μ = \frac{\sin ((A + δ_{m}) / 2)}{\sin (A / 2)}$ to find numerical $δ_{m}$ .

Execute: $\sin ((60^{\circ} + δ_{m}) / 2) = \sqrt{3} \cdot \sin (30^{\circ}) = \sqrt{3} / 2$ . Thus $(60^{\circ} + δ_{m}) / 2 = 60^{\circ} ⟹ 60^{\circ} + δ_{m} = 120^{\circ} ⟹ δ_{m} = 60^{\circ}$ . Testing options for $A = 60^{\circ}$ : (0) $180 - 120 = 60^{\circ}$ , (1) $90 - 60 = 30^{\circ}$ , (2) $180 + 120 = 300^{\circ}$ , (3) $180 - 180 = 0^{\circ}$ . Only Option A gives $60^{\circ}$ .

Verify: Option A uniquely matches the computed value $60^{\circ}$ .

Angle of Minimum Deviation for a Prism with Cotangent Refractive Index

What feels right?

Using the prism formula $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (A / 2)}$ with $μ = \cot (A / 2) = \frac{\cos (A / 2)}{\sin (A / 2)}$ , equating $\sin (\frac{A + δ_{m}}{2}) = \cos (A / 2) = \sin (90^{\circ} - A / 2)$ gives $δ_{m} = 180^{\circ} - 2 A$ .

Why each option works or fails

Solution

01Given

02Find

03Strategise

04Execute

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Students also ask

Angle of Minimum Deviation for a Prism with Cotangent Refractive Index

Using the prism formula μ=sin⁡(A+δm2)sin⁡(A/2)μ=sin(A/2)sin(2A+δm​​)​ with μ=cot⁡(A/2)=cos⁡(A/2)sin⁡(A/2)μ=cot(A/2)=sin(A/2)cos(A/2)​, equating sin⁡(A+δm2)=cos⁡(A/2)=sin⁡(90∘−A/2)sin(2A+δm​​)=cos(A/2)=sin(90∘−A/2) gives δm=180∘−2Aδm​=180∘−2A.

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Using the prism formula $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (A / 2)}$ with $μ = \cot (A / 2) = \frac{\cos (A / 2)}{\sin (A / 2)}$ , equating $\sin (\frac{A + δ_{m}}{2}) = \cos (A / 2) = \sin (90^{\circ} - A / 2)$ gives $δ_{m} = 180^{\circ} - 2 A$ .

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