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JEE Main 2026 · 8 Apr · Shift 2, Physics question 42

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

A horizontal interface with μ₁ = 1 above and μ₂ = 1.5 below, and a vertical normal through O. Ray AO comes down from the upper left at angle α to the normal and continues as OB at angle β below.
Choose one option
Answer and worked solution

Official answer: A

Angles with the normal (the vertical):

  • sin⁡α=213\sin\alpha=\dfrac2{\sqrt{13}}.
  • sin⁡β=CC2+16\sin\beta=\dfrac C{\sqrt{C^2+16}}.

Snell's law.
213=1.5CC2+16 ⇒ 4C2+64=29.25C2 ⇒ C≈1.6.\frac2{\sqrt{13}}=1.5\frac C{\sqrt{C^2+16}}\ \Rightarrow\ 4C^2+64=29.25C^2\ \Rightarrow\ C\approx1.6.

(A)

Topics:Refraction at plane surfaces
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