Sodium light of wavelengths 650 nm and 655 nm is used to study diffraction at a single slit of aperture 0.5 mm. The distance between the slit and the screen is 2.0 m. The separation between the positions of the first maxima of the diffraction pattern obtained in the two cases is______ $times$ $10^{-5} mathrm{~m}$.

Sodium light of wavelengths <img alt="\mathrm{650 \mathrm{~nm} \; and\; 655 \mathrm{~nm}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B650%20%5Cmathrm%7B%7Enm%7D%20%5C%3B%20and

Sodium light of wavelengths $\mathrm{650 \mathrm{~nm} \; and\; 655 \mathrm{~nm}}$ is used to study diffraction at a single slit of aperture $\mathrm{0.5 \mathrm{~mm.}}$ The distance between the slit and the screen is $\mathrm{2.0 \mathrm{~m.}}$ The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is __________ $\mathrm{10^{-5} \mathrm{~m}}$ .
Option: 1
3

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

$\mathrm{b= 0.5\times 10^{-3}m= 5\times 10^{-4}m}$
$\mathrm{D= 2m}$

For maxima,
$\mathrm{\beta = \frac{\left ( 2n+1 \right )}{2}\pi= \frac{\pi b\sin \theta }{\lambda }}$
For First maxima, $\mathrm{n= 1}$
$\mathrm{\frac{3\pi}{2}= \frac{\pi b\sin \theta }{\lambda }}$
$\mathrm{\sin \theta= \frac{3\lambda}{2b}$

Since , $\mathrm{\lambda }$ is very small
$\mathrm{\sin \theta \simeq \tan \theta = \frac{3\lambda }{2b}= \frac{y}{D}}$
$\mathrm{y = \frac{3 }{2} \frac{\lambda }{b}D}$

$\mathrm{For\; \lambda _{1}= 650\, nm,}$
$\mathrm{y_{1}= \frac{3}{2}\times \frac{650\times 10^{-9}\times 2}{5\times 10^{-4}}}$
$\mathrm{y_{1}= 390\times 10^{-5}m}$

$\mathrm{For\, \lambda _{2}= 655\, nm}$
$\mathrm{y_{2}= \frac{3}{2}\times \frac{655\times 10^{-9}\times 2}{5\times 10^{-4}}}$
$\mathrm{y_{2}= \393\times 10^{-5}m}$

$\mathrm{\Delta y= y_{2}-y_{1}= 3\times 10^{-5}m}$