Suppose that you have a reflection diffraction grating with n = 140 lines per mi

Question

Suppose that you have a reflection diffraction grating with n= 140 lines per millimeter. Light from a sodium lamp passes through the grating and is diffracted onto a distant screen.

A) Two visible lines in the sodium spectrum have wavelengths 498 nm and 569 nm. What is the angular separation of the first maxima of these spectral lines generated by this diffraction grating?'

B) How wide does this grating need to be to allow you to resolve the two lines 589.00 and 589.59 nanometers, which are a well known pair of lines for sodium, in the second order (m=2)?

Explanation / Answer

Given that

The number of lines per mm is =140

Now the grating element is d =1/140 =7.142*10-3mm =7.142*10-6m

Wavelengths lamda =498*10-9m

                   lamda' =569*10-9m

Now condition for the 1st maximum is

dsintheta =lamda

and sintheta =lamda/d

Then theta1 =sin-1(lamda/d) =3.998degrees

Now the for the lamda'

        theta2 =sin-1(lamda'/d) =4.5695degrees

Therefore the angular separation is =theta2-theta1 =0.571degrees

b)

Now given that wavelengths lamda =589nm

                                        lamda' =589.59nm

And now for order m=2 we know that

lamda'/dlamda =nN

Now [(lamda+lamda')/2]/[(lamda'-lamda)]=2N

       589.295/0.59 =2N ====>N =499.4 then nearly 500

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