Two spectral lines at wavelengths $590, nm$ and $594, nm$ are to be resolved using the fourth order ( $m = 4$ ) spectrum of a diffraction grating whose line spacing is $1.0, μ m$ . Using Rayleigh’s criterion, find the minimum total number of illuminated slits $N$ required.

hard 4 marks Class 12 · Physics · Wave Optics

Answer

N_{min} = 592

The correct answer is $N_{min} = 592$ — "However this is the *number of illuminated lines per order*. For a grating, the total number of illuminated slits equals $m N$ , so $N_{total} = m N = 4 \times 37 = 148$ .

Step-by-step solution

"Rayleigh’s criterion for a grating: $R = m N = \frac{l amb d a}{D e l t a l amb d a}$ where $R$ is the resolving power.
"Take the mean wavelength $λ_{avg} = \frac{590 + 594}{2} = 592, nm$ and the difference $D e l t a l amb d a = 4, nm$ .
" $m N = \frac{592 , nm}{4 , nm} = 148$
"Divide by the diffraction order $m = 4$ : $N = \frac{148}{4} = 37$
"However this is the *number of illuminated lines per order*. For a grating, the total number of illuminated slits equals $m N$ , so $N_{total} = m N = 4 \times 37 = 148$ .

Common mistake: Confusing

N

with

m N

or using the wrong wavelength in the formula

Concept tested: Resolving power of a grating

Published 16 September 2026 · How this solution was prepared and checked

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Two spectral lines at wavelengths 590,nm and 594,nm are to be resolved using the fourth order (m=4) spectrum of a diffraction grating whose line spacing is 1.0,μm. Using Rayleigh’s criterion, find the minimum total number of illuminated slits N required.

Step-by-step solution

Two spectral lines at wavelengths $590, nm$ and $594, nm$ are to be resolved using the fourth order ( $m = 4$ ) spectrum of a diffraction grating whose line spacing is $1.0, μ m$ . Using Rayleigh’s criterion, find the minimum total number of illuminated slits $N$ required.