In a diffraction pattern due to single slit of width 'a', the first minimum is observed at an angle of…
In a diffraction pattern due to single slit of width 'a', the first minimum is observed at an angle of $30^{\circ}$ when the light of wavelength $5400 Å$ is incident on the slit. The first secondary maximum is observed at an angle of $\left(\sin 30^{\circ}=\frac{1}{2}\right)$
$\sin ^{-1}\left(\frac{3}{4}\right)$
$\sin ^{-1}\left(\frac{2}{3}\right)$
$\sin ^{-1}\left(\frac{1}{2}\right)$
$\sin ^{-1}\left(\frac{1}{4}\right)$
Solution
For $\mathrm{n}^{\text {th }}$ secondary minimum, path difference $=a \sin \theta_{\mathrm{n}}=\mathrm{n} \lambda$ For $\mathrm{n}^{\text {th }}$ secondary maximum, path difference $=a \sin \theta_n=(2 n+1) \frac{\lambda}{2}$
$\therefore \quad$ For $1^{\text {st }}$ minimum, a $\sin 30^{\circ}=\lambda$
For $2^{\text {nd }}$ maximum, a $\sin \theta_{\mathrm{n}}=(2+1) \frac{\lambda}{2}=\frac{3 \lambda}{2}$
$\therefore \quad$ Dividing equation (i) by equation (ii),
$\frac{\left(\frac{1}{2}\right)}{\sin \theta_n}=\frac{2}{3} \Rightarrow \theta_n=\sin ^{-1}\left(\frac{3}{4}\right)$
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