Question

Two wavelengths of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction taking place at a single slit of aperture $2\times {10}^{-4}$ m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases.

Answer 1

For maxima other than central maxima
$a.\theta =(n+\frac{1}{2})\lambda$ and $\theta =\frac{y}{D}$
$\therefore a.\frac{y}{D}=(n+\frac{1}{2})\lambda$
For light of wavelength ${\lambda }_1=590nm$
$2\times {10}^{-14}\times \frac{y_1}{1.5}\times 590\times {10}^{-9}$
$y_1=\frac{3}{2}\times \frac{590\times {10}^{-9}\times 1.5}{2\times {10}^{-4}}=6.64mm$
For light of wavelength = 596 nm
$2\times {10}^{-4}\times \frac{y_2}{1.5}=(1+\frac{1}{2})\times 596\times {10}^{-9}$
$\Rightarrow y_2=\frac{3}{2}\times \frac{596\times {10}^{-9}\times 1.5}{2\times {10}^{-4}}=6.705mm$
Separation between two positions of first maxima
$\Delta y=y_2-y_1=6.705-6.64=0.065mm$