Question

(a) In what way is different from each slit related to the interference pattern in a double slit experiment
(b) Two wavelength 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 1.5m, Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases.

Answer 1

(a) If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double-slit experiment is modified by diffraction from each of the two slit.
(b) Give that Wavelength of the light beam, ${\lambda }_1=590nm=5.9\times {10}^{-7}m$
Wavelength of the light beam, ${\lambda }_2=596nm=5.96\times {10}^{-7}m$
Distance of the slit from the screen $=D+1.5m$
Distance between the two slits, $a=2\times {10}^{-4}m$
For the first secondary maxima, $sin\theta =\frac{3{\lambda }_1}{2a}=\frac{x_i}{D}$
$X_1=\frac{3{\lambda }_{1}D}{2a}and{X}_2=\frac{3{\lambda }_{2}D}{2a}$
$X_1=\frac{3\times 590\times {10}^{-9}\times 15}{2\times 2\times {10}^{-4}},X_2=\frac{3\times 596\times {10}^{-9}\times 1.5}{2\times 2\times {10}^{-4}}$
$X_1=663.75\times {10}^{-5},X_2=670.50\times {10}^{-5}$
$\therefore$ Spacing between the position of first secondary maxima of two sodium lines
$X_2-X_1=6.75\times {10}^{-5}m=0.0675mm$