Radiation sources and wavelength selectors MCQs With Answer

Radiation sources and wavelength selectors MCQs With Answer

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Radiation sources and wavelength selectors form the backbone of spectroscopic instrumentation widely used in pharmaceutical analysis. This topic explains common radiation sources — deuterium and xenon lamps, tungsten‑halogen, LEDs and lasers — and examines wavelength selectors such as monochromators, diffraction gratings, prisms, interference filters, Fabry–Pérot etalons and Michelson interferometers. Emphasis is placed on instrumental parameters like spectral bandwidth, resolving power, dispersion, slit width, stray light, throughput and detector matching, and how these affect sensitivity, linearity and method validation for UV‑Vis, IR and NIR assays. Practical understanding enables correct component choice, troubleshooting and improved assay accuracy.

Now let’s test your knowledge with 30 MCQs on this topic.

Q1. Which radiation source is most commonly used for continuous UV emission in UV‑Vis spectrophotometers?

  • Deuterium lamp
  • Tungsten‑halogen lamp
  • Xenon flash lamp
  • LED

Correct Answer: Deuterium lamp

Q2. Which source is typically used for covering the visible and near‑infrared (VIS–NIR) region with continuous output?

  • Deuterium lamp
  • Tungsten‑halogen lamp
  • Mercury vapor lamp
  • Helium‑neon laser

Correct Answer: Tungsten‑halogen lamp

Q3. Which wavelength selector disperses light by diffraction from ruled grooves and is widely used in monochromators?

  • Prism
  • Interference filter
  • Diffraction grating
  • Fabry–Pérot etalon

Correct Answer: Diffraction grating

Q4. Which equation correctly describes the diffraction grating condition for constructive interference?

  • d(sin i − sin θ) = mλ
  • d(sin i + sin θ) = mλ
  • n1 sin θ1 = n2 sin θ2
  • E = hc/λ

Correct Answer: d(sin i + sin θ) = mλ

Q5. The resolving power R of a diffraction grating is best expressed by which relation?

  • R = λ/Δλ = mN (order × number of illuminated grooves)
  • R = d/λ (groove spacing divided by wavelength)
  • R = F/# (f‑number of the monochromator)
  • R = throughput × slit width

Correct Answer: R = λ/Δλ = mN (order × number of illuminated grooves)

Q6. Which monochromator configuration commonly uses two mirrors and a plane grating and is prevalent in many spectrometers?

  • Czerny–Turner monochromator
  • Ebert‑Fastie monochromator
  • Prism monochromator
  • Fabry–Pérot monochromator

Correct Answer: Czerny–Turner monochromator

Q7. What is the main benefit of using a double monochromator in spectrophotometry?

  • Wider spectral bandwidth
  • Reduced stray light and improved photometric range
  • Lower dispersion
  • Smaller physical size

Correct Answer: Reduced stray light and improved photometric range

Q8. How does increasing the entrance slit width of a monochromator affect performance?

  • Decreases throughput and increases resolution
  • Increases throughput and decreases resolution
  • No effect on throughput but increases wavelength accuracy
  • Eliminates stray light completely

Correct Answer: Increases throughput and decreases resolution

Q9. Spectral bandwidth (bandpass) of a monochromator is primarily determined by which factors?

  • Detector type and source stability
  • Slit width and reciprocal linear dispersion of the monochromator
  • Lamp lifetime and ambient temperature
  • Grating blaze angle only

Correct Answer: Slit width and reciprocal linear dispersion of the monochromator

Q10. What is the purpose of the blaze angle on a diffraction grating?

  • To prevent higher order diffraction
  • To concentrate efficiency at a specific blaze wavelength
  • To increase the physical strength of the grating
  • To scatter unwanted wavelengths

Correct Answer: To concentrate efficiency at a specific blaze wavelength

Q11. Why are order‑sorting filters used with diffraction gratings?

  • To increase dispersion
  • To block overlapping higher diffraction orders that cause spectral contamination
  • To focus light onto the detector
  • To cool the grating

Correct Answer: To block overlapping higher diffraction orders that cause spectral contamination

Q12. The Michelson interferometer, used in FTIR, works by:

  • Splitting the beam and varying optical path difference to produce an interferogram
  • Diffracting light through a ruled grating
  • Transmitting only a single wavelength via absorption
  • Reflecting light through a rotating prism

Correct Answer: Splitting the beam and varying optical path difference to produce an interferogram

Q13. One major advantage of Fourier‑transform spectrometers over scanning monochromators is known as the Fellgett advantage, which refers to:

  • Higher spectral resolution at the same acquisition speed
  • Improved signal‑to‑noise because all wavelengths are measured simultaneously (multiplex advantage)
  • Lower cost of detectors
  • Reduced stray light exclusively

Correct Answer: Improved signal‑to‑noise because all wavelengths are measured simultaneously (multiplex advantage)

Q14. Which detector type is most commonly used for very low light, single‑photon sensitive detection in UV‑Vis instruments?

  • Charge‑coupled device (CCD)
  • Photomultiplier tube (PMT)
  • Thermopile
  • Bolometer

Correct Answer: Photomultiplier tube (PMT)

Q15. What is the main practical advantage of using a CCD detector in a spectrometer?

  • Extreme single‑photon sensitivity like PMTs
  • Simultaneous multichannel detection across many wavelengths
  • No need for any wavelength selector
  • Superior performance in the thermal IR without cooling

Correct Answer: Simultaneous multichannel detection across many wavelengths

Q16. Which detector material is most suitable for sensitive detection in the NIR (near‑infrared) region?

  • Silicon photodiode
  • InGaAs photodiode
  • PMT with bialkali photocathode
  • MCT (HgCdTe) for visible region

Correct Answer: InGaAs photodiode

Q17. Stray light in a spectrophotometer primarily causes which analytical error?

  • Overestimation of absorbance at very high absorbance values
  • Underestimation of absorbance at high absorbance values (flattening of baseline)
  • Wavelength calibration shift only
  • Complete loss of signal at the detector

Correct Answer: Underestimation of absorbance at high absorbance values (flattening of baseline)

Q18. In spectral terms, “bandpass” or spectral bandwidth is most accurately defined as:

  • The wavelength of maximum transmittance
  • The full width at half maximum (FWHM) of the transmitted spectral feature
  • The difference between the first and second diffraction orders
  • The angular spread of the collimated beam

Correct Answer: The full width at half maximum (FWHM) of the transmitted spectral feature

Q19. A Fabry–Pérot etalon selects wavelength by:

  • Single refraction through a dispersive glass
  • Multiple‑beam interference between parallel reflective surfaces
  • Absorption by a dyed gel
  • Diffraction from a ruled grating

Correct Answer: Multiple‑beam interference between parallel reflective surfaces

Q20. The minimum resolvable wavelength difference Δλ of a spectrometer is related to resolving power R by which expression?

  • Δλ = R × λ
  • Δλ = λ / R
  • Δλ = R / λ
  • Δλ = λ × f‑number

Correct Answer: Δλ = λ / R

Q21. Narrowing the monochromator slit width will generally produce which outcome?

  • Lower resolution and higher signal
  • Higher resolution and lower signal
  • No change in resolution but lower noise
  • Increased stray light

Correct Answer: Higher resolution and lower signal

Q22. How does increasing the groove density (lines per mm) of a diffraction grating affect dispersion?

  • Dispersion decreases with higher groove density
  • Dispersion increases with higher groove density
  • Dispersion is independent of groove density
  • Dispersion is only affected by slit width

Correct Answer: Dispersion increases with higher groove density

Q23. Which monochromator design commonly uses a single concave grating to perform both collimation and focusing, reducing mirror count?

  • Czerny–Turner
  • Ebert‑Fastie
  • Prism monochromator
  • Double‑pass plane grating monochromator

Correct Answer: Ebert‑Fastie

Q24. Thin‑film interference filters achieve wavelength selectivity primarily by:

  • Absorbing unwanted wavelengths in a dye layer
  • Phase interference between multiple dielectric layers to transmit narrow bands
  • Diffraction from micro‑grooves
  • Mechanical rotation of prisms

Correct Answer: Phase interference between multiple dielectric layers to transmit narrow bands

Q25. A monochromator with a smaller f‑number (faster optics) typically results in which effect?

  • Lower throughput and decreased signal
  • Higher throughput and increased signal at the detector
  • Reduced spectral resolution inherently
  • Elimination of stray light

Correct Answer: Higher throughput and increased signal at the detector

Q26. The Jacquinot advantage of Fourier‑transform instruments refers to which benefit?

  • Higher wavelength accuracy due to gratings
  • Improved throughput (no entrance slit) compared to scanning instruments
  • Complete absence of detector noise
  • Automatic wavelength calibration

Correct Answer: Improved throughput (no entrance slit) compared to scanning instruments

Q27. Which of the following is commonly used to test and quantify stray light in a spectrophotometer?

  • Holmium oxide wavelength standard
  • Band‑stop filter with high absorbance (cut‑off filter) at the test wavelength
  • Thermocouple calibration
  • Neutral density filter only

Correct Answer: Band‑stop filter with high absorbance (cut‑off filter) at the test wavelength

Q28. Reciprocal linear dispersion of a monochromator is usually expressed in which units?

  • nm/s
  • nm/mm
  • mm/nm
  • Hz/μm

Correct Answer: nm/mm

Q29. For pulsed high‑intensity short‑duration broadband light useful in rapid scanning spectrometers, which lamp is typically used?

  • Deuterium lamp
  • Tungsten lamp
  • Xenon flash lamp
  • LED array

Correct Answer: Xenon flash lamp

Q30. In Fourier‑transform spectroscopy, applying an apodization function to the interferogram primarily:

  • Increases peak intensity without changing width
  • Reduces sidelobe artifacts at the expense of some spectral resolution (peak broadening)
  • Eliminates the need for a beamsplitter
  • Converts the spectrum from absorbance to transmittance

Correct Answer: Reduces sidelobe artifacts at the expense of some spectral resolution (peak broadening)

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