The figure shows a ray of light entering one end of an optical fiber at an angle

Question

The figure shows a ray of light  entering one end of an optical fiber at an angle of incidence ?i = 48.5

The figure shows a ray of light entering one end of an optical fiber at an angle of incidence ?i = 48.5 degree . The index of refraction of the fiber is 1.43. Find the angle ? the ray makes with the normal when it reaches the curved surface of the fiber. Calculate the critical angle ?c for the optical fiber. Explain how this indicates that the internal reflection from the curved surface is total.

Explanation / Answer

( a)

for refraction n1*sin(i1)=n2*sin(i2) ;

n1=1; i1=48.5; and n2=1.43

sin(i2)= sin(48.5)/1.43;

i2=31.5 degree

i2+theta=90degree so theta = 90 - i2 = 90 - 31.5 = 58.5 degree

  

critical angle (c ) = sin -1(n1/n2) but n1 =1

so c= sin-1(1/1.43)

c=44.37 degree

(b)

since the light is incident at an angle which is more than critical angle

58.5> 44.37=c

and as light is moving in a denser medium so it will sufer a total internal reflaction

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