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JEE MAIN 2025
29-01-2025 SHIFT-1
Question
At the interface between two materials having refractive indices ${n_1}$ and ${n_2}$ , the critical angle for reflection of an em wave is ${\theta _{1c}}$. The ${n_2}$ material is replaced by another material having refractive index ${n_3}$ such that the critical angle at the interface between ${n_1}$ and ${n_3}$ materials is ${\theta _{2C}}$. If ${n_3} > {n_2} > {n_1};\frac{{{n_2}}}{{{n_3}}} = \frac{2}{5}$ and $\sin {\theta _{2C}} - \sin {\theta _{1C}} = \frac{1}{2}$, then ${\theta _{1C}}$ is
[Note : This questions marked as BONUS by NTA]
Select the correct option:
A
${\sin ^{ - 1}}\left( {\frac{1}{{6{n_1}}}} \right)$
B
${\sin ^{ - 1}}\left( {\frac{1}{{3{n_1}}}} \right)$
C
${\sin ^{ - 1}}\left( {\frac{5}{{6{n_1}}}} \right)$
D
${\sin ^{ - 1}}\left( {\frac{2}{{3{n_1}}}} \right)$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Solution Image
Question Tags
JEE Main
Physics
Hard
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