Question

Light of wavelength 500 nm is incident normally on a diffraction grating. The third-order maximum of the diffraction pattern is observed at 32.0°.

(a) What is the number of rulings per centimeter for the grating?_________grooves/cm

(b) Determine the total number of primary maxima that can be observed in this situation.__________

Answer 1

a) Using relation

\(d=\frac{m \lambda}{\sin \theta}\)

where \(\mathrm{d}\) is slit separation

\(d=3^{*} 500^{*} 10^{\wedge}-9 / \sin 32=2.83^{*} 10^{\wedge}-6 \mathrm{~m}\)

Hence number of rulings per \(m=1 / d=353279\)

per \(\mathrm{cm}=3532\) grooves \(/ \mathrm{cm}\)

b) Now

\(\sin \theta=\frac{m \lambda}{d}\)

\(\sin \theta=\frac{m * 500 * 10^{-9}}{2.83 * 10^{-6}}=0.1767 \mathrm{~m}\)

Now \(\sin \theta \leq 1\)

So \(\mathrm{m}=5.66\)

Hence maximum value of \(m=5\)

Hence total number of primary maxima \(=2 \mathrm{~m}+1=\mathbf{1 1}\)