7.The visibility `V` of an interference pattern is :
(A) `V=(I_(min))/(I_(max))`
(B) `V=I_(max)+I_(min)`
(C) `V=(I_(max)+I_(min))/(I_(max)-I_(min))`
(D) `V=(I_(max)-I_(min))/(I_(max)+I_(min))`

Two coherent sources of intensity ratio beta interfere, then (I_(max)-I_(min))/(I_(max)+I_(min)) is

Two coherent sources of intensity ratio alpha interfere. In interference pattern (I_(max)-I_("min"))/(I_(max)-I_("min")) =

Two doherent sources of intensity ratio alpha interfere in interference pattern (I_(max)-I_(min))/(I_(max)+I_(min)) is equal to

Two coherent sources of intensity ratio alpha interface . In interference pattern (I_("max") - I_("min"))/(I_("max") + I_("min")) =

In interference, two individual amplitudes are 5 units and 3 units. Find (a) A_(max)/A_(min) (b) I_(max)/I_(min) .

In interference, I_(max)/I_(min) = alpha, find. (a) A_(max)/A_(min) (b) A_1/A_2 (c) I_1/I_2

In interference, I_(max)/I_(min) = alpha , find (a) A_(max)/A_(min) (b) A_1/A_2 (c) I_1/I_2 .

Two coherent sources of light of intensity ratio beta produce interference pattern. Prove that in the interferencepattern (I_(max) - I_(min))/(I_(max) + (I_(min))) = (2 sqrt beta)/(1 + beta) where I_(max) and I_(min) are maximum and mininum intensities in the resultant wave.

The intensity ratio of the two interfering beams of light is m. What is the value of I_("max")-I_("min") // I_("max")+I_("min") ?