Light ray incident along a vector AO(AO = 2i - 3j) emerges out along vector OB(OB = Ci - 4j) as shown in the figure below. The value of C is ______. (a) 1.6 (b) 0.16 (c) 11.6 (d) 16

Answer: (a). From the figure the interface is along x and the normal along y, so sin α = (2)/(√(13)) and sin β = (C)/(√(C² + 16)). Snell's law: 1 × (2)/(√(13)) = 1.5 × (C)/(√(C² + 16)). Square: 4(C² + 16) = 29.25 C² ⇒ 25.25 C² = 64. C² ≈ 2.53 ⇒ C ≈ 1.6, so the answer is (a).

Light ray incident along a vector AO(AO = 2i - 3j) emerges out along… | JEE Main 2026 (8 Apr, Shift 2)

Q6JEE Main 2026 · 8 APR · SHIFT 2Medium

Refraction, Glass Slab & Apparent Depth

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

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Answer: (a)

From the figure the interface is along $x$ and the normal along $y$ , so $\sin\alpha = \dfrac{2}{\sqrt{13}}$ and $\sin\beta = \dfrac{C}{\sqrt{C^2 + 16}}$ .
Snell's law: $1 \times \dfrac{2}{\sqrt{13}} = 1.5 \times \dfrac{C}{\sqrt{C^2 + 16}}$ .
Square: $4(C^2 + 16) = 29.25\,C^2 \Rightarrow 25.25\,C^2 = 64$ .
$C^2 \approx 2.53 \Rightarrow C \approx 1.6$ , so the answer is (a).

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