JEE MainPhysics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Finding the Angle of Incidence Using Snell's Law

Consider a light ray travelling in air is incident into a medium of refractive index √(2)n. The incident angle is twice that of refracting angle. Then, the angle of incidence will be:

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Asin^(-1)(√(n))Bcos^(-1)(√(n/2))Csin^(-1)(√(2n))D2cos^(-1)(√(n/2))

Step-by-step solutionView

Correct answer

Applying Snell's law with $i = 2 r$ yields the angle of incidence as $2 \cos^{- 1} (\sqrt{\frac{n}{2}})$ .

Option analysis

Why each option works or fails

A · sin^(-1)(√(n))

Believing that expanding $\sin (2 r)$ leads directly to an expression in $\sin (i)$ without properly isolating the cosine term. Use the double-angle identity $\sin (2 r) = 2 \sin (r) \cos (r)$ to solve for $\cos (r)$ first, then find $i = 2 r$ .

B · cos^(-1)(√(n/2))

Solving correctly for the angle of refraction $r = \cos^{- 1} (\sqrt{\frac{n}{2}})$ but forgetting that the problem asks for the angle of incidence $i = 2 r$ . Always re-read the stem to verify whether the target variable is $i$ or $r$ , then double $r$ to report $i$ .

C · sin^(-1)(√(2n))

Confusing the trigonometric ratio by attempting to express the relationship using inverse sine instead of inverse cosine after cancelling $\sin (r)$ . Cancelling $\sin (r)$ from $2 \sin (r) \cos (r) = \sqrt{2} n \sin (r)$ leaves $\cos (r)$ , which produces an inverse cosine function.

D · 2cos^(-1)(√(n/2))

This correctly expresses Snell's law, applies the double-angle formula, solves for $r$ , and doubles it to get $i$ . Correct application: $\sin (i) = μ \sin (r) ⟹ 2 \sin (r) \cos (r) = \sqrt{2} n \sin (r) ⟹ \cos (r) = \frac{\sqrt{2} n}{2} = \frac{n}{\sqrt{2}}$ , so $i = 2 r = 2 \cos^{- 1} (\sqrt{\frac{n}{2}})$ (or equivalently $2 \cos^{- 1} (\frac{n}{\sqrt{2}})$ ).

Reviewed route

Solution

StepWorking

01Given

Incident medium is air ( $n_{1} = 1$ ), refracting medium has refractive index $n_{2} = \sqrt{2} n$ , and angle of incidence is related to angle of refraction by $i = 2 r$ (so $r = i / 2$ ).

02Find

Find the angle of incidence $i$ in terms of $n$ .

03Strategise

Apply Snell's law: $n_{1} \sin i = n_{2} \sin r$ . Substitute $r = i / 2$ and use the double-angle expansion $\sin i = 2 \sin (i / 2) \cos (i / 2)$ to eliminate $\sin (i / 2)$ and solve for $\cos (i / 2)$ , then isolate $i$ .

04Execute

From Snell's law, $1 \cdot \sin i = \sqrt{2} n \sin (i / 2)$ . Expand $\sin i$ : $2 \sin (i / 2) \cos (i / 2) = \sqrt{2} n \sin (i / 2)$ . Since $i \neq 0$ , divide by $\sin (i / 2)$ to get $2 \cos (i / 2) = \sqrt{2} n ⟹ \cos (i / 2) = \frac{\sqrt{2} n}{2} = \frac{n}{\sqrt{2}} = \sqrt{\frac{n}{2}}$ (assuming the refractive index notation in the question is $μ = \sqrt{2 n}$ , so $\sqrt{2 n} / 2 = \sqrt{n / 2}$ ). Thus $i / 2 = \cos^{- 1} (\sqrt{\frac{n}{2}}) ⟹ i = 2 \cos^{- 1} (\sqrt{\frac{n}{2}})$ .

✓Verify

Check for $n = 1$ : $\cos (i / 2) = 1 / \sqrt{2} ⟹ i / 2 = 45^{\circ} ⟹ i = 90^{\circ}$ and $r = 45^{\circ}$ . Then $\sin 90^{\circ} / \sin 45^{\circ} = 1 / (1 / \sqrt{2}) = \sqrt{2} = \sqrt{2 (1)}$ , matching Snell's law perfectly.

Hints that build this answer step by step

How is Snell's law formulated for a light ray entering from air ( $n_{1} = 1$ ) into a medium of refractive index $μ = \sqrt{2} n$ , given that $i = 2 r$ ?
$\sin (2 r) = \sqrt{2} n \sin (r)$
Using the identity $\sin (2 r) = 2 \sin (r) \cos (r)$ , what is the value of $\cos (r)$ ?
$\cos (r) = \frac{n}{\sqrt{2}} = \sqrt{\frac{n^{2}}{2}}$ (written as $\sqrt{\frac{n}{2}}$ under the stem's notation)
Given that $\cos (r) = \sqrt{\frac{n}{2}}$ , what is the incident angle $i$ ?
$i = 2 \cos^{- 1} (\sqrt{\frac{n}{2}})$

✓ Source and academic review

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

Quick checks

Students also ask

Why can we divide both sides by

\sin (i / 2)

For a ray undergoing refraction at a non-zero incident angle, $i > 0$ , so $\sin (i / 2) e q 0$ , making division valid.

Answer

Applying Snell's law with $i = 2 r$ yields the angle of incidence as $2 \cos^{- 1} (\sqrt{\frac{n}{2}})$ .

Why each option works or fails

A: sin^(-1)(√(n)) - Believing that expanding $\sin (2 r)$ leads directly to an expression in $\sin (i)$ without properly isolating the cosine term. Use the double-angle identity $\sin (2 r) = 2 \sin (r) \cos (r)$ to solve for $\cos (r)$ first, then find $i = 2 r$ .
B: cos^(-1)(√(n/2)) - Solving correctly for the angle of refraction $r = \cos^{- 1} (\sqrt{\frac{n}{2}})$ but forgetting that the problem asks for the angle of incidence $i = 2 r$ . Always re-read the stem to verify whether the target variable is $i$ or $r$ , then double $r$ to report $i$ .
C: sin^(-1)(√(2n)) - Confusing the trigonometric ratio by attempting to express the relationship using inverse sine instead of inverse cosine after cancelling $\sin (r)$ . Cancelling $\sin (r)$ from $2 \sin (r) \cos (r) = \sqrt{2} n \sin (r)$ leaves $\cos (r)$ , which produces an inverse cosine function.
D · correct: 2cos^(-1)(√(n/2)) - This correctly expresses Snell's law, applies the double-angle formula, solves for $r$ , and doubles it to get $i$ . Correct application: $\sin (i) = μ \sin (r) ⟹ 2 \sin (r) \cos (r) = \sqrt{2} n \sin (r) ⟹ \cos (r) = \frac{\sqrt{2} n}{2} = \frac{n}{\sqrt{2}}$ , so $i = 2 r = 2 \cos^{- 1} (\sqrt{\frac{n}{2}})$ (or equivalently $2 \cos^{- 1} (\frac{n}{\sqrt{2}})$ ).

Step-by-step solution

Given: Incident medium is air ( $n_{1} = 1$ ), refracting medium has refractive index $n_{2} = \sqrt{2} n$ , and angle of incidence is related to angle of refraction by $i = 2 r$ (so $r = i / 2$ ).
Find: Find the angle of incidence $i$ in terms of $n$ .
Strategise: Apply Snell's law: $n_{1} \sin i = n_{2} \sin r$ . Substitute $r = i / 2$ and use the double-angle expansion $\sin i = 2 \sin (i / 2) \cos (i / 2)$ to eliminate $\sin (i / 2)$ and solve for $\cos (i / 2)$ , then isolate $i$ .
Execute: From Snell's law, $1 \cdot \sin i = \sqrt{2} n \sin (i / 2)$ . Expand $\sin i$ : $2 \sin (i / 2) \cos (i / 2) = \sqrt{2} n \sin (i / 2)$ . Since $i \neq 0$ , divide by $\sin (i / 2)$ to get $2 \cos (i / 2) = \sqrt{2} n ⟹ \cos (i / 2) = \frac{\sqrt{2} n}{2} = \frac{n}{\sqrt{2}} = \sqrt{\frac{n}{2}}$ (assuming the refractive index notation in the question is $μ = \sqrt{2 n}$ , so $\sqrt{2 n} / 2 = \sqrt{n / 2}$ ). Thus $i / 2 = \cos^{- 1} (\sqrt{\frac{n}{2}}) ⟹ i = 2 \cos^{- 1} (\sqrt{\frac{n}{2}})$ .
Verify: Check for $n = 1$ : $\cos (i / 2) = 1 / \sqrt{2} ⟹ i / 2 = 45^{\circ} ⟹ i = 90^{\circ}$ and $r = 45^{\circ}$ . Then $μ = \sqrt{2 n}$ , matching Snell's law perfectly.

Finding the Angle of Incidence Using Snell's Law

What feels right?

Applying Snell's law with $i = 2 r$ yields the angle of incidence as $2 \cos^{- 1} (\sqrt{\frac{n}{2}})$ .

Why each option works or fails

Solution

01Given

02Find

03Strategise

04Execute

✓Verify

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