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Tell me? - Optics - JEE Main

In figure, the optical fiber is $l=2\; m$ long and has a diameter of $d=20\; \mu m.$ If ray of light is incident on one end of the fiber at angle $\Theta _{1}=40^{o},$ the number of reflections it makes before emerging from the other end close to :

(refractive index of fiber is $1.31$ and $\sin 40^{o}=0.64$ )

Option 1)
$66000$
Option 2)
$45000$
Option 3)
$55000$
Option 4)
$57000$

Answers (1)

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S solutionqc

$l=2m$

$d=20\; \mu m$

$\Theta _{1}=40^{o}$

$\mu =1.31$

Snells law for 1st surface

$1.\sin (40)=\mu .\sin r=1.31\sin r=0.64$

$\sin r=\frac{0.64}{1.31}\approx \frac{1}{2}$

$\Rightarrow \: r=30^{o}$

and $\Theta =90-r=60^{o}$

For T.I.R at 2nd surface

$\Theta >C$ and $(\sin \Theta >\sin C\Rightarrow \sin 60>\frac{1}{1.3}\Rightarrow \frac{\sqrt{3}}{2}>\frac{1}{1.3})$

So $\Theta >C$

So T.I.R will happen

So $\tan r=\frac{d}{l_{1}}\Rightarrow l_{1}=\frac{d}{\tan (30)}=20\sqrt{3}\mu m$

$n=no.of \; reflection$

So $n=\frac{l}{l_{1}}=\frac{2}{20\sqrt{3}\times 10^{-6}}$

$n\approx 57000$

Option 1)

$66000$

Option 2)

$45000$

Option 3)

$55000$

Option 4)
$57000$

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