Fixed Czerny–Turner Spectrograph Coverage and Resolution
An undergraduate team is building a Czerny–Turner spectrograph for 400–700 nm. A 1200 lines/mm grating is used in first order with a 100 mm focal-length collimator and a 100 mm focal-length camera mirror. The entrance slit is 50 µm wide. The detector is a 2048-pixel linear CCD with 12 µm pixels, and the team wants to cover 400–700 nm in a single acquisition without moving the grating. The source is a broadband continuum with a strong 532 nm laser line that must not appear in the wrong order. The mechanical drawing shows the grating can be rotated, but the team wants a fixed configuration. A labmate claims the 12 µm pixels give 0.1 nm resolution because 'one pixel equals 0.1 nm'.
1. Calculate the linear dispersion at 500 nm and the spectral coverage of the 2048-pixel detector. Does the design meet the 400–700 nm requirement?
Read the worked answer
The grating constant is d = 1/(1200 mm⁻¹) = 833.3 nm. At 500 nm in first order, the grating equation gives sinα + sinβ = λ/d = 500/833.3 = 0.600. For a near-symmetric Czerny–Turner, take α ≈ β, so sinβ ≈ 0.300 and cosβ ≈ 0.954.
The linear dispersion is dx/dλ = mf/(d cosβ) = (1)(100 × 10⁻³ m)/[(833.3 × 10⁻⁹ m)(0.954)] ≈ 1.26 × 10⁵. Thus 1 nm maps to about 0.126 mm, or 126 µm. Equivalently, one 12 µm pixel spans about 12/126 = 0.095 nm.
The detector width is 2048 × 12 µm = 24.576 mm. The covered range is therefore 24.576 mm/(0.126 mm/nm) ≈ 195 nm. If the short-wavelength end is set to 400 nm, the long-wavelength end is only about 595 nm. The design does not cover 400–700 nm in one fixed acquisition.
To cover 300 nm with this detector, the dispersion must be reduced to about 24.576/300 = 0.082 mm/nm, roughly 82 µm/nm. That would require a lower line density, a shorter focal length, or a larger detector. A 600 lines/mm grating, for example, roughly doubles the coverage but also roughly doubles the slit-limited spectral resolution.
2. What is the actual spectral resolution set by the 50 µm slit? How many pixels sample the slit image, and is the labmate's 0.1 nm claim correct?
Read the worked answer
The slit image on the detector has width w_image = w_slit (f₂/f₁) = 50 µm × (100/100) = 50 µm. With the dispersion above, the spectral width corresponding to that image is Δλ_slit = 50 µm/(126 µm/nm) ≈ 0.40 nm.
The number of pixels sampling the slit image is 50 µm/12 µm ≈ 4.2 pixels. That satisfies Nyquist sampling of the slit image, but it does not make the resolution equal to one pixel. The resolution is slit-limited at about 0.4 nm, not 0.1 nm.
The labmate's statement confuses sampling interval with resolving power. A 0.1 nm resolution would require a slit image of about 12.6 µm, which is only about one pixel wide and would be undersampled by this CCD. Even with a narrower slit, the pixel pitch would limit the sampled resolution to roughly 0.19 nm for a two-pixel sampling criterion. To reach 0.1 nm, the team needs higher dispersion, a narrower slit, and a detector with smaller pixels or more pixels, not just a reinterpretation of the existing pixel size.
3. The 532 nm line is present. Identify any order-overlap risk in the 400–700 nm range and recommend an order-sorting solution. Also state one change that would let the same detector cover 400–700 nm.
Read the worked answer
With the grating used in first order over 400–700 nm, the second order of 200–350 nm falls at the same angles as the first-order 400–700 nm light. If the continuum source has any UV output, those UV wavelengths will overlap the visible spectrum. The 532 nm line itself is not the main second-order problem in this window: its second order is at 266 nm, outside the displayed range, and its third order is at 1593 nm. However, a strong 532 nm line can still cause stray light, ghosts, and detector blooming, and any UV-generated second-order light can contaminate the visible spectrum.
Use an order-sorting filter: a 400 nm long-pass filter before the slit removes the UV that would appear in second order. If the 532 nm line is not the signal of interest, add a 532 nm notch or edge filter, but be aware that it also removes part of the continuum near 532 nm. A wedged or tilted filter avoids etalon fringes.
To cover 400–700 nm with the same 2048 × 12 µm detector, reduce the dispersion to about 82 µm/nm. A practical change is to replace the 1200 lines/mm grating with roughly an 800 lines/mm grating, or to reduce the camera focal length. Either change increases coverage but degrades slit-limited resolution; the team should state which specification is more important and verify the resulting resolution with the 50 µm slit.