JEE Main 2025 Jan 29 Shift 1, Physics Q8: Critical Angle and Total Internal Reflection

JEE Main2025Jan 29, Shift 1Physics

At the interface between two materials having refractive indices $n_{1}$ and $n_{2}$ , the critical angle for total internal reflection is $θ_{1 C} .$ The $n_{2}$ material is replaced by another material of refractive index $n_{3}$ , such that the critical angle at the $n_{1}$ – $n_{3}$ interface is $θ_{2 C} .$ If $n_{2} > n_{3} > n_{1}$ , $\frac{n _{3}}{n _{2}} = \frac{2}{5}$ , and $sin θ_{2 C} - sin θ_{1 C} = \frac{1}{2}$ , then $θ_{1 C}$ is

A
$sin^{- 1} (\frac{1}{6})$
B
$sin^{- 1} (\frac{2}{3})$
C
$sin^{- 1} (\frac{5}{6})$
D
$sin^{- 1} (\frac{1}{3})$

Solution

TIR requires the ray to travel from the denser block into the rarer outside medium, so $n_{2}, n_{3} > n_{1} .$ With $n_{2} > n_{3}$ the critical angles are

$sin θ_{1 C} = \frac{n _{1}}{n _{2}}, sin θ_{2 C} = \frac{n _{1}}{n _{3}},$

and $sin θ_{2 C} > sin θ_{1 C}$ , consistent with the given positive difference. Using $\frac{n _{3}}{n _{2}} = \frac{2}{5} :$

$sin θ_{2 C} - sin θ_{1 C} = \frac{n _{1}}{n _{2}} (\frac{n _{2}}{n _{3}} - 1) = \frac{n _{1}}{n _{2}} \cdot \frac{3}{2} = \frac{1}{2} .$

$∴ \frac{n _{1}}{n _{2}} = \frac{1}{3} ⟹ sin θ_{1 C} = \frac{1}{3} ⟹ θ_{1 C} = sin^{- 1} (\frac{1}{3}) .$

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JEE Main 2025 Jan 29 Shift 1, Physics Q8: Critical Angle and Total Internal Reflection

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