Question

A miniature spectrometer used for chemical analysis has a diffraction grating with 800 slits/mm set 25.0 mm in front of the detector "screen." The detector can barely distinguish two bright lines that are 30 µm apart in the first-order spectrum. What is the resolution of the spectrometer at a wavelength of 600 nm? That is, if two distinct wavelengths can barely be distinguished, one of them being 600.0 nm, what is the wavelength difference Δλ between the two?

Answer 1

Δλ = 13.16 nm

Step-by-step explanation:

Given:

- Diffraction grating d = 1 / 800 = 0.00125 mm

- Focal length of the lens f = 25.0 mm

- The difference of bright lines Δ x = 30 um

- wavelength of light λ = 600 nm

Find:

what is the wavelength difference Δλ between the two?

Solution:

- The separation between the bright lines is proportional to sin(Q) as follows:

x = f*sin(Q)

- Since the bright lines are very close we can use small angle approximations:

sin(Q) ≈ Q

- The relation above becomes:

x = f*Q

- The differential form can be written as:

Δx = f*ΔQ

ΔQ = Δx/f

- Compute ΔQ: ΔQ = 30*10^-6 / 25*10^-3 = 0.0012 rads

- We know from the Young's experiment that the primary maxima occurs at:

λ = d*sin(Q) / m

- Given the lights in first order of the spectrum, hence, m = 1:

λ = d*sin(Q)

- Differentiate the above expression with respect to angle Q

Δλ = d*cos(Q)*ΔQ

- replace cos(Q) = sqrt ( 1 - sin^2(Q)) and sin(Q) = λ / d:

Δλ = d*sqrt ( 1 - ( λ / d)^2)*ΔQ

- Plug in the values:

Δλ = (0.00125*10^-3)*sqrt ( 1 - ( 600*10^-9 / 0.00125*10^-3)^2)*0.012

Δλ = (0.00125*10^-3)*0.8773*0.012

Δλ = 13.16 nm