Two wavelength 590 nm and 596 nm of sodium light are used one after other, to study the diffraction taking…
Two wavelength 590 nm and 596 nm of sodium light are used one after other, to study the diffraction taking place at a single slit of aperture 2.4 mm . The distance between the slit and screen is 2 m . The separation between the positions of first secondary maximum of the diffraction pattern obtained in the two cases is
$7.5 \times 10^{-6} \mathrm{~m}$
$7.5 \times 10^{-9} \mathrm{~m}$
$2.5 \times 10^{-6} \mathrm{~m}$
$5.0 \times 10^{-6} \mathrm{~m}$
Solution
The first secondary maximum in a single slit diffraction pattern is obtained at
$\sin \theta=(2 \mathrm{n}+1) \frac{\lambda}{2 \mathrm{a}}=\frac{3 \lambda}{2 \mathrm{a}}$
for small angles, $\sin \theta \approx \tan \theta=\theta$
$\begin{array}{ll}\therefore & \frac{3 \lambda}{2 a}=\frac{x}{D} \Rightarrow x=\frac{3 \lambda D}{2 a} \\ \therefore & \Delta x=\frac{3 D \times\left(\lambda_2-\lambda_1\right)}{2 d}=\frac{3 \times 2 \times(596-590) \times 10^{-9}}{2 \times 2.4 \times 10^{-3}} \\ \therefore & \Delta x=7.5 \times 10^{-6} m\end{array}$