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The speed of light in two media $c_1$ and $c_2$ are $1.5 \times 10^8 \mathrm{~m} / \mathrm{s}$ and $2 \times 10^8 \mathrm{~m} / \mathrm{s}$ respectively. If the light undergoes total internal reflection, the critical angle between the two media is
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The correct answer is:
$\sin ^{-1}\left(\frac{3}{4}\right)$
The refractive index of the medium is defined as, $\mu_m=\frac{c}{c_m}$, i.e., $c_m$ is the speed of light in medium and $c$ is the speed of light in vaccum.
Since, $c_1\mu_2$ and medium 2 is the rarerer medium.
According to snells law,
$\mu_m \sin \theta_m=\mu \sin \theta=$ constant
At critical angle, the angle of incidence in denser medium is $\theta_{\mathrm{C}}$ critical and angle of refraction in rarerer medium is $\theta=\frac{\pi}{2}$.
Therefore,
$\begin{aligned} & \mu_1 \sin \theta_{1 C}=\mu_2 \sin \left(\frac{\pi}{2}\right) \\ & \Rightarrow \sin \theta_{1 \mathrm{C}}=\left(\frac{\mu_2}{\mu_1}\right)=\left(\frac{c_1}{c_2}\right)=\frac{1.5 \times 10^8 \mathrm{~m} / \mathrm{s}}{2 \times 10^8 \mathrm{~m} / \mathrm{s}}=\frac{3}{4} \\ & \Rightarrow \theta_{1 \mathrm{C}}=\sin ^{-1}\left(\frac{3}{4}\right)\end{aligned}$
Since, $c_1
According to snells law,
$\mu_m \sin \theta_m=\mu \sin \theta=$ constant
At critical angle, the angle of incidence in denser medium is $\theta_{\mathrm{C}}$ critical and angle of refraction in rarerer medium is $\theta=\frac{\pi}{2}$.
Therefore,
$\begin{aligned} & \mu_1 \sin \theta_{1 C}=\mu_2 \sin \left(\frac{\pi}{2}\right) \\ & \Rightarrow \sin \theta_{1 \mathrm{C}}=\left(\frac{\mu_2}{\mu_1}\right)=\left(\frac{c_1}{c_2}\right)=\frac{1.5 \times 10^8 \mathrm{~m} / \mathrm{s}}{2 \times 10^8 \mathrm{~m} / \mathrm{s}}=\frac{3}{4} \\ & \Rightarrow \theta_{1 \mathrm{C}}=\sin ^{-1}\left(\frac{3}{4}\right)\end{aligned}$
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