Light ray incident along a vector $\overrightarrow{\mathrm{AO}}(\overrightarrow{\mathrm{AO}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}})$ emerges out along vector $\overrightarrow{\mathrm{OB}}(\overrightarrow{\mathrm{OB}}=\mathrm{C} \hat{\mathrm{i}}-4 \hat{\mathrm{j}})$ as shown in the figure below. The value of C is ______.

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Geometrical Optics

Light ray incident along a vector

AO (AO = 2 \hat{i} - 3 \hat{j})

emerges out along vector

OB (OB = C \hat{i} - 4 \hat{j})

as shown in the figure below. The value of C is ______.

(A)1.6✓ Correct
(B)0.16
(C)11.6
(D)16

Answer: (A)

Solution

The interface is horizontal with its normal along

\hat{j}

, so the

\hat{i}

component of each ray is tangential. For the incident ray

A O = 2 \hat{i} - 3 \hat{j}

sin θ_{1} = \frac{2}{13}

; for the refracted ray

O B = C \hat{i} - 4 \hat{j}

sin θ_{2} = \frac{C}{C ^{2} + 16}

. Snell's law

μ_{1} sin θ_{1} = μ_{2} sin θ_{2}

gives

\frac{2}{13} = 1.5 \cdot \frac{C}{C ^{2} + 16}

. Squaring,

\frac{4}{13} = \frac{2.25 C ^{2}}{C ^{2} + 16}

, so

4 (C^{2} + 16) = 29.25 C^{2}

and

C^{2} = \frac{64}{25.25} = 2.53

. The value of

C

1.6

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option

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