The diffraction pattern of a light of wavelength $$400 \text{ nm}$$ diffracting from a slit of width $$0.2 \text{ mm}$$ is focused on the focal plane of a convex lens of focal length $$100 \text{ cm}$$. The width of the $$1^{st}$$ secondary maxima will be :

In single slit diffraction, the positions of the minima are given by:

$$a \sin\theta = n\lambda$$

where $$a$$ is the slit width, $$\lambda$$ is the wavelength, and $$n$$ is the order of the minimum.

For small angles, $$\sin\theta \approx \tan\theta = \frac{y}{f}$$, where $$y$$ is the position on the focal plane and $$f$$ is the focal length of the lens.

The position of the $$n$$-th minimum is:

$$y_n = \frac{n\lambda f}{a}$$

The width of the 1st secondary maximum is the distance between the 1st and 2nd minima:

$$\Delta y = y_2 - y_1 = \frac{2\lambda f}{a} - \frac{\lambda f}{a} = \frac{\lambda f}{a}$$

Substituting the given values: $$\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}$$, $$f = 100 \text{ cm} = 1 \text{ m}$$, $$a = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}$$:

$$\Delta y = \frac{400 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = \frac{400 \times 10^{-9}}{2 \times 10^{-4}} = 2 \times 10^{-3} \text{ m} = 2 \text{ mm}$$

The correct answer is $$2 \text{ mm}$$.