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Ray Optics and Optical Instruments — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsRay Optics and Optical Instruments2 Marks
Question
A point source is placed at a depth h below the surface of water (refractive index = μ).
  1. Show that light escapes through a circular area on the water surface with its centre directly above the point source.
  2. Find the angle subtended by a radius of the area on the source.
✓

Answer

  1. Let, x = radius of the circular area
$\frac{\text{x}}{\text{h}}=\tan\theta_\text{C}$ (where C is the critical angle)

$\Rightarrow\frac{\text{x}}{\text{h}}=\frac{\sin\theta_\text{C}}{\sqrt{1-\sin^2\theta_\text{C}}}=\frac{\frac{1}{\mu}}{\sqrt{1-\frac{1}{\mu^2}}} \ \Big(\because \ \sin\theta_\text{C}=\frac{1}{\mu}\Big)$

$\Rightarrow\frac{\text{x}}{\text{h}}=\frac{1}{\sqrt{\mu^2-1}}$ or $\text{x}=\frac{\text{h}}{\sqrt{\mu^2-1}}$

So, light escapes through a circular area on the water surface directly above the point source.
  1. Angle subtained by a radius of the area on the source, $\theta_\text{C}=\sin^{-1}\Big(\frac{1}{\mu}\Big)$

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