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)

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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