KCET · Physics · Ray Optics
The critical angle for a monochromatic light going from medium A to medium B is \(\theta\). If the speed of light in medium A is \(V\), then the speed of light in medium B is
- A \(V(1 - \cos\theta)\)
- B \(\dfrac{V}{\cos\theta}\)
- C \(\dfrac{V}{\sin\theta}\)
- D \(V(1 - \sin\theta)\)
Answer & Solution
Correct Answer
(C) \(\dfrac{V}{\sin\theta}\)
Step-by-step Solution
Detailed explanation
Using Snell's law for the critical angle when light travels from medium A to medium B:
\(\mu_A \sin\theta = \mu_B \sin 90^\circ\)
\(\sin\theta = \dfrac{\mu_B}{\mu_A}\)
Since the refractive index of a medium is inversely proportional to the speed of light in that medium, \(\mu = \dfrac{c}{v}\)
\(\dfrac{\mu_B}{\mu_A} = \dfrac{v_A}{v_B}\)
Given the speed of light in medium A is \(v_A = V\)
\(\sin\theta = \dfrac{V}{v_B}\)
\(v_B = \dfrac{V}{\sin\theta}\)
Answer: \(\dfrac{V}{\sin\theta}\)
\(\mu_A \sin\theta = \mu_B \sin 90^\circ\)
\(\sin\theta = \dfrac{\mu_B}{\mu_A}\)
Since the refractive index of a medium is inversely proportional to the speed of light in that medium, \(\mu = \dfrac{c}{v}\)
\(\dfrac{\mu_B}{\mu_A} = \dfrac{v_A}{v_B}\)
Given the speed of light in medium A is \(v_A = V\)
\(\sin\theta = \dfrac{V}{v_B}\)
\(v_B = \dfrac{V}{\sin\theta}\)
Answer: \(\dfrac{V}{\sin\theta}\)
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