A typical laboratory diffraction grating has 4800 lines/cm , and these lines are

Question

A typical laboratory diffraction grating has 4800 lines/cm , and these lines are contained in a 3.60 cm width of grating.

Part A

What is the chromatic resolving power of such a grating in the first order?

Part B

Could this grating resolve the lines of the sodium doublet (1 = 589.00nm, 2 = 589.59nm ) in the first order?

Part C

While doing spectral analysis of a star, you are using this grating in the second order to resolve spectral lines that are very close to the 587.8002-nm spectral line of iron. For wavelengths longer than the iron line, what is the shortest wavelength you could distinguish from the iron line?

Part D

For wavelengths shorter than the iron line, what is the longest wavelength you could distinguish from the iron line?

Part E

What is the range of wavelengths you could not distinguish from the iron line?(Values of shorter and longer you have calculated in two previous parts.)

A typical laboratory diffraction grating has 4800 lines/cm, and these lines are contained in a 3.60 cm width of grating. What is the chromatic resolving power of such a grating in the first order? R = Could this grating resolve the lines of the sodium doublet (lambda_1 = 589.00mn, lambda_2 = 589.59 nm) in the first order? yes no While doing spectral analysis of a star, you are using this grating in the second order to resolve spectral lines that are very close the line 587.8002-nm spectral line of iron. For wavelengths longer than the iron line, what is the shortest wavelength you could distinguish from the iron line? For wavelengths shorter than the iron line, what is the longest wavelength you could distinguish from the iron line? What is the range of wavelengths you could not distinguish from the iron line? (values of have calculated in two previous parts.)

Explanation / Answer

a) R = N m = 4800 * 3.6 * 1 = 17,280

b) The resolving power = lambda / delta lambda = 589 / (589.59 - 589) = 998

yes grating can easily resolver the doublet

c) R = lambda / delta lambda, R = 17,280 when m = 1

when m = 2, R = 2 * 17280 = 34,560

delta lambda = 587.8 / 34560 = 0.017 nm

lambdamin = lambda + delta lambda = 587.8002 + 0.017 = 587.82 nm

d)lambdamax = lambda - delta lambda = 587.8002 - 0.017 = 587.78 nm

e) Range = (587.78 nm, 587.82 nm)

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