00:01
Hello students in this question we have regional by prism which is used to obtain fringes from a point of source point source which is placed at a distance as equals to 2 meter and prism is between midway between the source and screen so and the wavelength lambda not it is equals to 500 nanometer and index of refraction small n is 1 .5 so we have to determine the prism angle we have to determine the prism angle that that is alpha, prism angle alpha, we have to determine here if the separation between the fringe is delta y, it is equal to 0 .5 m.
00:41
So we know that the position delta y, it is equal to lemma, s multiplied by lambda, 0, divided by 2d, alpha and n minus, sorry, the, sorry, this formula.
00:59
So we know that the position, we can write that.
01:02
That the a it is equals to 2d modplied by n minus 1 multiplied by alpha.
01:08
And alpha from here we can rearrange this equation to get s lambda s lambda divided by 2d and n minus 1 multiplied by this delta y.
01:24
So this a is having the another values which has been rearranged as we know that the position delta y it is equal to to s by a multiplied by lambda.
01:36
So this equation has been used.
01:38
It's in place of alpha.
01:40
So we have obtained this angle alpha.
01:42
So we can substitute the values.
01:44
So we have s and we don't have d...