00:01
Hi students in this question, the fraction by grating is being discussed and we need to find out the width of first order rainbow which is taken as delta y.
00:09
And in order to find the value of delta y, a few data are being given which include the number of grating capital letter n to be equal to 600 lines per millimeter.
00:20
Now from the value of this n we can find out the length of the grating small letter d to be equal to 1 divided by capital letter n.
00:28
Now by substituting the value of n in this equation, we get the value of length of the grating small letter d to be equal to 1 ,667 nanometer.
00:38
Now the condition which is used for explaining the diffraction process is taken as m into lambda will be equal to d into sine theta.
00:47
Let this be equation number one.
00:49
And here since we are dealing with the first order, the value of m will be equal to a value of 1.
00:53
Now we can say that when a light source which is having a wavelength lambda to be equal to 400 nanometer is being used, in that case we can write the value of theta 1 to be equal to sine inverse the value of lambda divided by d.
01:10
Now here while substituting the value of lambda and d in this equation, we get theta 1 to be equal to sine inverse the value of 400 divided by 1 ,667, which on simplification gives us.
01:23
Value of theta 1 to be equal to 13 .884 degree.
01:29
On the other hand, when a light source which is having a wavelength lambda to be equal to 700 nanometer is being used, in that case, the value of theta 2 will be equal to sine inverse the value of lambda divided by d...
