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JEE MAIN 2026 · Optics

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 42

Updated
Single correct+4 / -1Answer matches NTA’s final key

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

Ray AO in the medium of index 1 meets the boundary at O and refracts along OB into the medium of index 1.5; alpha and beta are measured from the normal

Choose one

Solution

Solution

The boundary is horizontal, so the normal is along j^\hat j and each angle's sine is the horizontal component over the length.

  • Incidence: sin⁡α=24+9=213\sin\alpha = \dfrac{2}{\sqrt{4 + 9}} = \dfrac{2}{\sqrt{13}}
  • Refraction: sin⁡β=CC2+16\sin\beta = \dfrac{C}{\sqrt{C^2 + 16}}

Snell's law, 1⋅sin⁡α=1.5sin⁡β1 \cdot \sin\alpha = 1.5\sin\beta:

413=2.25 C2C2+16  ⇒  4C2+64=29.25 C2  ⇒  C2=6425.25≈2.53,\frac{4}{13} = \frac{2.25\,C^2}{C^2 + 16} \;\Rightarrow\; 4C^2 + 64 = 29.25\,C^2 \;\Rightarrow\; C^2 = \frac{64}{25.25} \approx 2.53,

so C≈1.59≈1.6C \approx 1.59 \approx 1.6.

Answer: 1.6.

Answer checked against the official answer key. The solution is our own working.

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