00:01
For this problem on the topic of interference, we are shown a figure in which the waves along rays 1 and 2 are initially in phase, and they have the same wavelength lambda and a air.
00:10
Ray 2 goes to a material with length l and a refractive index of n, and the rays are then reflected onto a common point p on a screen by a mirror.
00:21
Now, if we can vary l from 0 to 2 ,400 nanometers, and the intensity of light at point p varies with l, as given in the figure, we want to know at what values of l is the intensity maximum zero, and then we want to find the multiple of the wavelength that will give the phase difference between ray 1 and 2 at the common point p when l is 1 ,200 nanometers.
00:48
Now when the interference between the two waves is completely destructive, the phase difference phi is equal to 2m plus 1 times pi where m is 0 or any positive integer value.
01:08
Now the equivalent condition is that their part length difference is an odd multiple of lambda over 2 where lambda is the wavelength of light.
01:17
And so looking at the figure we see that half of the periodic pattern is of length delta l, which is 750 nanometers, judging from the maximum at x is equal to zero to the minimum at x is 750 nanometers...