Ray Optics

JEE Main level, single correct, medium

89 of 295

Question 89 of 295

The critical angle for a denser-rarer interface is 45∘45^{\circ}. The speed of light in rarer medium is 3×108ms−13\times {10}^{8}m{s}^{-1}. The speed of light in the denser medium is:

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Answer: (C) 2.12×108ms−12.12\times {10}^{8}m{s}^{-1}

The critical angle and the refractive index of a media are related by the formula:
sin⁡θc=(1μ)...(1)\sin {\theta }_{c}=\left(\dfrac{1}{\mu }\right)...\left(1\right)
Substitute the value of the critical angle into equation (1) and solve to calculate the refractive index of the medium.
sin⁡45∘=1μ⇒12=1μ⇒μ=2\begin{matrix}\sin 45^{\circ} & = & \dfrac{1}{\mu } \\ & \Rightarrow & \dfrac{1}{\sqrt{2}}=\dfrac{1}{\mu } \\ & \Rightarrow & \mu =\sqrt{2}\end{matrix}
The refractive index of the medium can also be written as
μ=cv...(2)\mu =\dfrac{c}{v}...\left(2\right)
Substitute the values of the known parameters into equation (2) and solve to calculate the required speed of light within the medium.
2=3×108ms−1v⇒v=3×108ms−12=2.12×108ms−1\begin{matrix}\sqrt{2} & = & \dfrac{3\times {10}^{8}m{s}^{-1}}{v} \\ & \Rightarrow & v=\dfrac{3\times {10}^{8}m{s}^{-1}}{\sqrt{2}} \\ & = & 2.12\times {10}^{8}m{s}^{-1}\end{matrix}

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