A semi-cylinder made of transparent plastic has a refractive index  n = $\sqrt{2}$ and a radius of R. There is a narrow incident light ray perpendicular to the flat side of the semi cylinder at d distance from the axis of symmetry.

What can the maximum value of d be so that the light ray can still leave the other side of the semi cylinder?

• A

  $\mathrm{d}=\frac{\mathrm{R}}{\sqrt{2}}$

• B

  $\mathrm{d}=\frac{\mathrm{R}}{2\sqrt{2}}$

• C

  $\frac{\mathrm{R}}{2}$

• D

  None of these

When the value of d chosen is such that TIR just takes place then time for which light remains inside the cylinder is:

• A

  $\frac{4\mathrm{R}}{\mathrm{c}}$

• B

  $\frac{4\sqrt{2}\mathrm{R}}{\mathrm{c}}$

• C

  $\frac{2\sqrt{2}\mathrm{R}}{\mathrm{c}}$

• D

  $\frac{2\mathrm{R}}{\mathrm{c}}$

Distance d is now varied, so that the ray always emerges from the other side of the semi cylinder. What is the range of OB?

• A

  $2\sqrt{2}\mathrm{R}>\mathrm{OB}>\mathrm{R}\sqrt{2}$

• B

  $5\sqrt{2}\mathrm{R}>\mathrm{OB}>\mathrm{R}$

• C

  $\mathrm{\infty }>\mathrm{OB}>\mathrm{R}\sqrt{2}$

• D

  $\mathrm{\infty }>\mathrm{OB}>2\mathrm{R}\sqrt{2}$

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