In Young's double slit experiment, carried out with light of wavelength $$5000 \text{ \AA}$$, the distance between the slits is $$0.3 \text{ mm}$$ and the screen is at $$200 \text{ cm}$$ from the slits. The central maximum is at $$x = 0 \text{ cm}$$. The value of $$x$$ for third maxima is ______ mm.

We are given that $$\lambda = 5000$$ Å $$= 5 \times 10^{-7}$$ m, $$d = 0.3$$ mm $$= 3 \times 10^{-4}$$ m, and $$D = 200$$ cm $$= 2$$ m.

The position of the $$n$$-th maximum in YDSE is given by

$$x_n = \frac{n\lambda D}{d}$$

For the third maximum ($$n = 3$$), we then have

$$x_3 = \frac{3 \times 5 \times 10^{-7} \times 2}{3 \times 10^{-4}} = \frac{30 \times 10^{-7}}{3 \times 10^{-4}} = 10 \times 10^{-3} = 10 \text{ mm}$$

Hence, the third maximum is located at 10 mm from the central maximum.