00:01
Here in this question, we know that the light through the film, light should undergo destructive interference condition.
00:11
And for destructive interference in the film, we can express the relation as 2 multiply by n, multiply by t equal to m multiply by lambda.
00:27
Now here n is the refractive index of the material, t is the thickness, m is the order of the fringe, and lambda is the wavelength.
00:39
So now comes to part a of this question.
00:43
So here we are going to find out the minimum thickness of the film that will cancel the light of wavelength 505 nanometer.
00:55
So here to calculate the minimum thickness, of the film we are using the above relation so for this we know that for minimum thickness the order of frames that is m must be equal to one so now we substitute 2 .62 for n 1 for m and 505 nanometer for lambda in the equation 2 multiply by n multiplied by t equal m multiply by lambda so we get 2 multiplied by 2 .62 multiply by t equals to 1 multiplied by 505 nanometer so now let us rearrange this for t so we get t equals to 1 multiplied by 500 5 nanometer so now let us rearrange this for t so we get t equals to 5 505 divided by 2 multiply by 2 .62.
02:20
So we get t equals to 96 .4 nanometer.
02:29
Thus, the minimum thickness of the film that will cancel light of wavelength 505 nanometer is equals to 96 .4 nanometer.
02:41
Now comes to part p of this question.
02:45
So here, to calculate the minimum thickness of the film for the three thinnest cases, we again use the relation to multiply by n, multiply by t equal to m multiply by lambda...