JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 12)
In figure, the optical fiber is $$l$$ = 2m long and
has a diameter of d = 20 μm. If a ray of light
is incident on one end of the fiber at angle
$$\theta _1$$ = 40°, the number of reflection it makes
before emerging from the other end is close to:
(refractive index of fibre is 1.31 and
sin 40° = 0.64)
_8th_April_Morning_Slot_en_12_1.png)
57000
55000
66000
45000
Explanation
If we approximate the angle $$\theta _2$$ as 30° initially then answer will be closer to 57000. but if
we solve thoroughly, answer will be close to 55000.
So both the answers must be awarded. Detailed solution as following.
EXACT SOLUTION
By Snell’s law 1.sin 40° = (1.31) sin $$\theta _2$$
$$\sin {\theta _2} = {{0.64} \over {1.31}} = {{64} \over {131}} \approx .49$$
Now tan$$\theta _2$$
= $${{64} \over {\sqrt {{{\left( {131} \right)}^2} - {{\left( {64} \right)}^2}} }} = {{64} \over {\sqrt {13065} }} \approx {{64} \over {114.3}} = {d \over x}$$
Now number of reflections
$$ = {{2 \times 64} \over {114.3 \times 20 \times {{10}^{ - 6}}}} = {{64 \times {{10}^5}} \over {114.3}}$$
$$ \approx 55991 \approx 55000$$
APPROXIMATE SOLUTION
By Snell’s law 1.sin 40° = (1.31)sin $$\theta _2$$
$$\sin {\theta _2} = {{0.64} \over {1.31}} = {{64} \over {131}} \approx .49$$
If assume $$ \Rightarrow $$ $$\theta _2$$ $$\approx $$ 30°
tan 30° = $${d \over x} \Rightarrow x = {\sqrt3} d$$
Now number of reflections
$$ = {t \over {\sqrt 3 d}} = {2 \over {\sqrt 3 \times 20 \times {{10}^{ - 6}}}} = {{{{10}^5}} \over {\sqrt 3 }}$$
$$ \approx 57735 \approx 57000$$
So both the answers must be awarded. Detailed solution as following.
EXACT SOLUTION
By Snell’s law 1.sin 40° = (1.31) sin $$\theta _2$$
$$\sin {\theta _2} = {{0.64} \over {1.31}} = {{64} \over {131}} \approx .49$$
Now tan$$\theta _2$$
= $${{64} \over {\sqrt {{{\left( {131} \right)}^2} - {{\left( {64} \right)}^2}} }} = {{64} \over {\sqrt {13065} }} \approx {{64} \over {114.3}} = {d \over x}$$
Now number of reflections
$$ = {{2 \times 64} \over {114.3 \times 20 \times {{10}^{ - 6}}}} = {{64 \times {{10}^5}} \over {114.3}}$$
$$ \approx 55991 \approx 55000$$
APPROXIMATE SOLUTION
By Snell’s law 1.sin 40° = (1.31)sin $$\theta _2$$
$$\sin {\theta _2} = {{0.64} \over {1.31}} = {{64} \over {131}} \approx .49$$
If assume $$ \Rightarrow $$ $$\theta _2$$ $$\approx $$ 30°
tan 30° = $${d \over x} \Rightarrow x = {\sqrt3} d$$
Now number of reflections
$$ = {t \over {\sqrt 3 d}} = {2 \over {\sqrt 3 \times 20 \times {{10}^{ - 6}}}} = {{{{10}^5}} \over {\sqrt 3 }}$$
$$ \approx 57735 \approx 57000$$
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