Critical angle of incidence for a pair of optical media is \(45^\circ\).
The refractive indices of the first and second media are in the ratio:
The refractive indices of the first and second media are in the ratio:
- \(\sqrt{2} : 1\)
- \(1 : 2\)
- \(1 : \sqrt{2}\)
- \(2 : 1\)
The Correct Option is A
Approach Solution - 1
To solve this question, we need to find the ratio of the refractive indices of the two media given the critical angle of incidence. The critical angle (\( \theta_c \)) for a pair of optical media is the angle of incidence in the denser medium for which the angle of refraction is \( 90^\circ \) in the rarer medium. Using Snell's Law, which states:
\(n_1 \sin \theta_c = n_2 \sin 90^\circ\)
Where:
- \( n_1 \) = refractive index of the denser medium
- \( n_2 \) = refractive index of the rarer medium
- \(\theta_c\) = critical angle
Given that the critical angle \( \theta_c = 45^\circ \), and \(\sin 90^\circ = 1\), the equation becomes:
\(n_1 \sin 45^\circ = n_2\)
We know \(\sin 45^\circ = \frac{\sqrt{2}}{2}\). Therefore:
\(n_1 \frac{\sqrt{2}}{2} = n_2\)
Rearrange to find the ratio of refractive indices:
\(\frac{n_1}{n_2} = \sqrt{2}\)
Hence, the refractive indices ratio is \( \sqrt{2} : 1 \). This matches the correct given option \(\sqrt{2} : 1\).
Let's analyze why the other options are incorrect:
- Option: \(1:2\) — This implies a small critical angle, not \(45^\circ\).
- Option: \(1:\sqrt{2}\) — This implies the media are reversed with \(\sqrt{2}:1\), which could not yield a critical angle of \(45^\circ\).
- Option: \(2:1\) — This also implies a smaller critical angle, inconsistent with \(45^\circ\).
Approach Solution -2
The critical angle is given by:
\[ \sin C = \frac{n_2}{n_1}. \]
At \( C = 45^\circ \):
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{n_2}{n_1}. \]
Thus:
\[ \frac{n_1}{n_2} = \sqrt{2} : 1. \]
Final Answer: \(\sqrt{2} : 1\).
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