MHT CET · Physics · Ray Optics
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
- A \(\sin ^{-1}\left(\frac{2}{3}\right)\)
- B \(\sin ^{-1}\left(\frac{4}{3}\right)\)
- C \(\sin ^{-1}\left(\frac{3}{2}\right)\)
- D \(\sin ^{-1}\left(\frac{3}{4}\right)\)
Answer & Solution
Correct Answer
(D) \(\sin ^{-1}\left(\frac{3}{4}\right)\)
Step-by-step Solution
Detailed explanation
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,
\(\mu_1 \sin \theta_{1 C}=\mu_2 \sin \left(\frac{\pi}{2}\right)\)
\(\Rightarrow \sin \theta_{1 \mathrm{C}}=\left(\frac{\mu_2}{\mu_1}\right)=\left(\frac{c_1}{c_2}\right)=\) \(\frac{1.5 \times 10^8 \mathrm{~m} / \mathrm{s}}{2 \times 10^8 \mathrm{~m} / \mathrm{s}}=\frac{3}{4}\)
\(\Rightarrow \theta_{1 \mathrm{C}}=\sin ^{-1}\left(\frac{3}{4}\right)\)
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,
\(\mu_1 \sin \theta_{1 C}=\mu_2 \sin \left(\frac{\pi}{2}\right)\)
\(\Rightarrow \sin \theta_{1 \mathrm{C}}=\left(\frac{\mu_2}{\mu_1}\right)=\left(\frac{c_1}{c_2}\right)=\) \(\frac{1.5 \times 10^8 \mathrm{~m} / \mathrm{s}}{2 \times 10^8 \mathrm{~m} / \mathrm{s}}=\frac{3}{4}\)
\(\Rightarrow \theta_{1 \mathrm{C}}=\sin ^{-1}\left(\frac{3}{4}\right)\)
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