Suppose that you have a reflection diffraction grating with n=125 lines per mill

Question

Suppose that you have a reflection diffraction grating with n=125 lines per millimeter. Light from a sodium lamp passes through the grating and is diffracted onto a distant screen.

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?

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

n=125 lines per millimeter

1 = 498nm

2 = 569nm

maxima produced is

dsin=m

d = 1/n = 1/125 = 8*10^-6 m

sin = m1/d

= sin^-1(498*10^-9 /8*10^-6 m )

= 3.569 deg

dispersion of grating = D = m/dcos

D = 1/(8*10^-6)*( cos3.569)

D = 125242.9017

Angular separation

=D

= 125242.9017 (71*10^-9)

= 8.892*10^-3

b)

Resolving power = R = / = 589/.59 = 998.3

Width of grating = R/m = 998.3/2 = 499

width w = 499/125 mm^-1

= 3.992 mm

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