Derive the expression for refractive index of the material of the prism in terms of angle of the prism and angle of minimum deviation.

Explanation

From `Delta MQR, ( i-r_(1) + ( e-r_(2)) = delta`
so `(i+e) - ( r_(1) + r_(2)) = delta `
From `Delta PQN , r_(1) + r_(2) + /_QNR = 180^(@)`
Also `A + /_ QNR = 180^(@)`
Thus `A = r_(1) + r_(2)`
So `i+e - A = delta `
At minimum deviation,` i=e,r_(1) = r_(2) = r ` and `delta = delta_(m)`
`rArr i = ( A + delta_(m))/( 2)`
and `r = ( A) /( 2)`
Also `mu = ( sin i)/(sinr )`
Hence, `mu = ( sin ((A+ delta_(m))/(2)))/(sin((A)/(2)))`
(b)
Each optical fibre consists of a core and cladding. Refractive index of the material of the core is higher than that of the cladding. When a signal, in the form of light, is directed into the optical fibre, at an angle greater than the (relevant) critical angle, it undergoes repeated total internal reflections along the length of the fibre and comes out of it at the other end with almost negligible loss of intensible.