Fraunhofer single-slit diffraction condition

For a slit of width $a$, the minima occur at $a\sin\theta = m\lambda, \quad m=1,2,3,\dots$

On the focal plane of a lens of focal length $f$, for small angles, $y \approx f\tan\theta \approx f\sin\theta$ so the position of the $m^{\text{th}}$ minimum is $y_m = \frac{m\lambda f}{a}.$

Width of the first secondary maximum

The first secondary maximum lies between the first and second minima. Hence its width is the distance between these minima: $\text{width} = y_2 - y_1 = \frac{2\lambda f}{a} - \frac{\lambda f}{a} = \frac{\lambda f}{a}.$

Substitute the given values

Given: $\lambda = 400\,\text{nm} = 4\times 10^{-7}\,\text{m}$ $a = 0.2\,\text{mm} = 2\times 10^{-4}\,\text{m}$ $f = 100\,\text{cm} = 1\,\text{m}$

Therefore, $\text{width} = \frac{\lambda f}{a} = \frac{4\times 10^{-7}\times 1}{2\times 10^{-4}} = 2\times 10^{-3}\,\text{m}.$

Converting to mm: $2\times 10^{-3}\,\text{m} = 2\,\text{mm}.$

Check options

• A: $2\,\text{mm}$ ✔
• B: $0.2\,\text{mm}$ ✘
• C: $0.02\,\text{mm}$ ✘
• D: $2\,\text{cm}$ ✘

Final answer

The width of the first secondary maximum is $\boxed{2\,\text{mm}}$ so the correct option is A.