Resolvance of Diffraction Gratings

In order to know if a diffraction grating is able to resolve two wavelengths, the resolving power of the diffraction grating must be found
- This is based on the Rayleigh criterion as applied to diffraction gratings and their output
The resolving power, R, of a diffraction grating is given by:

R = \frac{λ}{Δ λ}

Where:
- R = the resolving power of the grating (no unit)
- λ = the wavelength of incident light (m)
- Δλ = the smallest difference in wavelength that the grating can resolve (m)

The resolving power is also equal to

R = N × m

Where:
- N = the total number of slits on the diffraction grating (or those illuminated by an incident beam of light)
- m = the order of diffraction
Therefore, combining the two equations gives:

R = \frac{λ}{Δ λ} = N \times m

Worked example

A student is using a diffraction grating to resolve two emission wavelengths from calcium in the 3^rd order of the spectrum. These wavelengths are 164.99 nm and 165.20 nm.

Determine the minimum number of lines per mm needed if a beam of width 0.25 mm is incident upon the diffraction grating.

Step 1: List the known values

- Order of diffraction, m = 3
- Wavelength, λ₁ = 164.99 nm
- Wavelength, λ₂ = 165.20 nm
- Beam width = 0.25 mm

Step 2: Determine values for λ and Δλ

- The value of the incident wavelength, λ, can be determined from the mean of the two wavelengths:

$λ = \frac{164.99 + 165.20}{2}$ = 165.095 nm

- The difference between the two wavelengths, Δλ, is:

Δλ = λ₂ − λ₁= 165.20 − 164.99 = 0.21 nm

Step 3: Find the resolving power of these wavelengths

- The resolving power can be calculated using:

R = \frac{λ}{Δ λ}

Step 4: Input the relevant values

R = \frac{λ}{Δ λ} = \frac{165.095}{0.21}

= 786.2

Step 5: Find the number of lines illuminated

- The equation relating resolving power and number of slits is given by:

R = N × m

- Rearranging for N and substituting the values for R and m:

$N = \frac{R}{m} = \frac{786.2}{3}$ = 262.1

- The value 262.1 is the number of lines illuminated for the beam of 0.25 mm

Step 6: Find the number of lines needed per mm for this situation

- Since 262.1 is the number of lines illuminated for the beam of 0.25 mm, then 4 times more lines must be illuminated to account for 1 mm:

262.1 × 4 = 1048.4 lines per mm

State 7: State the final answer

- The minimum amount of lines needed per mm to resolve the relevant calcium lines is: 1049 lines per mm

Examiner Tip

In the worked example, the answer may look as though it has been rounded incorrectly but we are looking for the actual number of lines here, not fractions of lines

Rounding down to 1048 would leave the grating nearly half a line short, which can't happen, so always round up to the nearest whole number

Resolvance of Diffraction Gratings (DP IB Physics: HL)

Revision Note