An optical fiber has index of refraction `n=1.40` and diameter `d=100 mu m`. It is surrounded by air. Light is sent into the fiber along the axis as shown in figure. If smallest outside radius R permitted for a bend in the fiber for no light to escape is given by `50 x ("in" mu m)` fill value of x.

Answer

7

Explanation

A ray along the inner edge will escape. Its angle of incidence is described by `sin theta = (R-d)/(R )` and by n `sin theta gt sin `90^(@)`
Then `(n(R-d))/(R ) gt 1 nR - nd gt R nR -R gt nd R gt (nd)/(n-1)`
`R_(min) =(1.40(100xx10^(-6)m))/(0.40) =350xx 10^(-6)m`