A light ray in the core ( n = 1.45) of a cylindrical optical fiber (the figure b

Question

A light ray in the core (n = 1.45) of a cylindrical optical fiber (the figure below) is incident on the cladding. A ray is transmitted through the cladding (n = 1.20) and into the air. The emerging ray makes an angle theta2 = 8.50



A light ray in the core (n = 1.45) of a cylindrical optical fiber (the figure below) is incident on the cladding. A ray is transmitted through the cladding (n = 1.20) and into the air. The emerging ray makes an angle theta2 = 8.50 A light ray in the core (n = 1.45) of a cylindrical optical fiber (the figure below) is incident on the cladding. A ray is transmitted through the cladding (n = 1.20) and into the air. The emerging ray makes an angle theta2 = 8.50 A light ray in the core (n = 1.45) of a cylindrical optical fiber (the figure below) is incident on the cladding. A ray is transmitted through the cladding (n = 1.20) and into the air. The emerging ray makes an angle theta2 = 8.50

Explanation / Answer

we have,

sin(i)/sin(r) = n2/n1

for cladding-air refraction

sin(theta'2)/sin(90-8.50) = ng/nc

sin(theta'2)= 1/1.20*sin(90-8.50)

==> theta'2 = arcsin(1/1.20*sin(90-8.50))

==> theta'2 = 55.5054322 degree

so,

for core-cladding refraction

sin(theta'1)/sin(theta'2) = nc/nf

sin(theta'1)= 1.20/1.45*sin(55.5054322 degree)

==> theta'1 = arcsin(1.20/1.45*sin(55.5054322 degree) )

==> theta'1 = 43.0063888 degree

theta1 = 90-theta'1 = 90-43.0063888 degree = 46.99 degree = 47 degree

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