1. Condition for minima in single-slit diffraction

For a single slit of width $a$, the minima are given by

$a\sin\theta = m\lambda, \qquad m=1,2,3,\dots$

For small angles,

$\sin\theta \approx \tan\theta \approx \frac{y}{D}$

So the position of the $m$-th minimum on the screen is

$y_m = \frac{m\lambda D}{a}$

2. Distance between first and third minima

First minimum:

$y_1 = \frac{\lambda D}{a}$

Third minimum:

$y_3 = \frac{3\lambda D}{a}$

Hence the distance between them is

$y_3-y_1 = \frac{3\lambda D}{a}-\frac{\lambda D}{a} = \frac{2\lambda D}{a}$

Given:

• Wavelength: $6000\,\text{\AA} = 6000\times 10^{-10}\,\text{m} = 6\times 10^{-7}\,\text{m}$
• Screen distance: $D=50\,\text{cm}=0.5\,\text{m}$
• Distance between first and third minima: $3\,\text{mm}=3\times 10^{-3}\,\text{m}$

So,

$\frac{2\lambda D}{a} = 3\times 10^{-3}$

$a = \frac{2\lambda D}{3\times 10^{-3}}$

3. Substitute values

$a = \frac{2\times 6\times 10^{-7}\times 0.5}{3\times 10^{-3}}$

$a = \frac{6\times 10^{-7}}{3\times 10^{-3}}$

$a = 2\times 10^{-4}\,\text{m}$

4. Required integer

The slit width is

$2\times 10^{-4}\,\text{m}$

So the blank is:

$\boxed{2}$