Question

Paragraph for Question 1 and 2
A semi cylinder made of a transparent plastic has a refraction index of j = V2 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.

1. 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= R/√2
(B) d=R/2√2
(C)d=R/2
(D) None of these

2. When the value of d chosen is such that TIR just takes place then time for which light remains
inside the cylinder is :- (c = speed of light in vacuum)

(A)4R/c
(B)4√2R/c
(C)2√2R/c
(D)2R/c



Experts please solve the problem....

Answer 1

Answer:

ur name tiyaaa

Explanation:

my name is also tiya

Answer 2

Option A)$d=\frac{R}{\sqrt{2}}$ and Option C) $\frac{2\sqrt{2}R}{c}$ are the answers.

Given:
Refractive index of plastic $=\sqrt{2}$

Radius = R

Distance from the axis of symmetry = d

Speed of light in vacuum = c

To find:

The maximum value of d so light can still leave the other side of the semi-cylinder.

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

Solution:

At the maximum value of d, the angle of incidence at the curved surface will be equal to the critical angle and hence the light will go tangential to the curved surface after leaving the semi-cylinder.

The angle of incidence $\theta = \sin^{-1}\frac{d}{R}$.

Therefore using the refraction formula at the curved surface

$\sqrt{2} \sin\theta = 1* \sin90$

$\sqrt{2}*\frac{d}{R} = 1$

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

• When TIR takes place, $d=\frac{R}{\sqrt{2}}$ and $\theta = 45$.
• Therefore the light will reflect at an angle 90 to the original incident ray and hence vertically downwards.
• Again the incident angle is 45 and the light will reflect at an angle of 90 to this incident light and go back horizontally at a distance d below the axis of symmetry.
• Hence the total distance travelled by the light is 4d

Hence the time taken to travel a 4d distance

$t = \frac{4d}{c} =\frac{2\sqrt{2}R }{c}$

The Maximum value of d so ray can still leave the other side $= \frac{R}{\sqrt{2}}$.

Time for which light remains inside the cylinder if TIR takes place$= \frac{2\sqrt{2}R}{c}$

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