First slide

Spherical refracting surface's

Question

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.

Moderate

Question

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

A

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

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

$\frac{R}{2}$
D

None of these

Solution

$\begin{array}{l} \sin θ_{C} = \frac{1}{\sqrt{2}} \\ \Rightarrow θ_{C} = 45^{\circ} \Rightarrow d = \frac{R}{\sqrt{2}} \end{array}$

Question

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

A

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

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

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

$\frac{2 R}{c}$

Solution

Total distance travel in the semi cylinder $= 2 \sqrt{2} R$
Optical path travel $= (2 \sqrt{2} R) \times \sqrt{2} = 4 R$
$time = \frac{4 R}{c}$

Question

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

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

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

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

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

Solution

$Range R \sqrt{2} to \infty$

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