± A Diffraction Grating Spectrometer: Suppose that you have a reflection diffrac

Question

± A Diffraction Grating Spectrometer:

Suppose that you have a reflection diffraction grating with n= 115 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

part a:

slit width=d=1 mm/115=8.6957*10^(-6) m

for a diffraction grating:

m*lambda=d*sin(theta)

where m=order of the spectral line

theta=angle of diffraction

lambda=wavelength

for m=1:

lambda=498 nm

==>theta1=asin(lambda/d)=3.2831 degrees

lambda=569 nm

==>theta2=3.7518 degrees

angular separation=theta2-theta1=0.46872 degrees

part b:

resolving power=wavelength/difference of wavelength=N*m

where N=number of slits illuminated

hence 589/(589.59-589)=N*2

==>N=499.15

so width=499.15/115=4.34 mm

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