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# calculate the critical angle and an angle of entry for an optical fibre having core of refractive index 1.50 and cladding of refractive index 1.48a. 30 degree b. 14.2 degreec. 5 deg d. 45 degree

Khimraj
3007 Points
3 years ago
Let n1 = refractive index of core = 1.50
n2 = refractive index of cladding = 1.48
critical angle is given by
sin$\Theta _{c}$ = n2/n1
So $\Theta _{c}$ = sin-1(1.48/1.50) = 80.63o
and angle of entry is given by
sin$\alpha$ = $\sqrt{n_{1}^{2} - n_{2}^{2}}$
$\alpha$ = 14.2o