JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
The critical angle for a denser-rarer interface is \(45^{\circ}\). The speed of light in rarer medium is \(3 \times 10^8\) ms. The speed of light in the denser medium is:
- A \(5 \times 10^7\,m / s\)
- B \(2.12 \times 10^8\,m / s\)
- C \(3.12 \times 10^7\,m / s\)
- D \(\sqrt{2} \times 10^8\,m / s\)
Answer & Solution
Correct Answer
(B) \(2.12 \times 10^8\,m / s\)
Step-by-step Solution
Detailed explanation
\(i _{ C }=\) Critical angle \(\frac{v}{C}=\frac{1}{\mu}=\sin i_C=\sin 45^{\circ}=\frac{1}{\sqrt{2}}\) \(\Rightarrow v=\frac{ C }{\sqrt{2}}=\frac{3 \times 10^8}{\sqrt{2}}\,m / s =2.12 \times 10^8\,m / s\)
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