In a Young's double slit setup, the two slits are separated by a distance $d = 1.5 \mathrm{~mm}$. Wh

Question

In a Young's double slit setup, the two slits are separated by a distance $d = 1.5 \mathrm{~mm}$. What should be the width of each individual slit ($a$) so that exactly 12 interference maxima are contained within the central diffraction maximum of the single slit pattern?

Professor PC Thomas Classes · Answer

(a) : The angular width of the central diffraction maximum is $	heta_c = \frac{2\lambda}{a}$. The angular separation between consecutive interference maxima is $	heta_i = \frac{\lambda}{d}$. The number of interference fringes within the central maximum is $n = \frac{	heta_c}{	heta_i} = \frac{2\lambda / a}{\lambda / d} = \frac{2d}{a}$. Given $n = 12$, we have $12 = \frac{2d}{a} \implies a = \frac{2d}{12} = \frac{d}{6}$. Substituting $d = 1.5 \mathrm{~mm}$, we get $a = \frac{1.5}{6} = 0.25 \mathrm{~mm}$.