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

The correct answer is 10.

Given Data:
Wavelength of light, $\lambda =5000\text{Å}=5000\times {10}^{-10}\text{m}=5\times {10}^{-7}\text{m}$
Distance between the slits, $d=0.3\text{mm}=0.3\times {10}^{-3}\text{m}=3\times {10}^{-4}\text{m}$
Distance of the screen from the slits, $D=200\text{cm}=2\text{m}$
Order of maximum, $n=3$ (for third maxima)

Formula:
The position of the $n^{\text{th}}$ maxima from the central maximum in a Young's double slit experiment is given by:
$x=\frac{n\lambda D}{d}$

Calculation:
Substitute the given values into the formula to find the position $x$ of the third maxima ($n=3$):
$x=\frac{3\times (5\times {10}^{-7}\text{m})\times 2\text{m}}{3\times {10}^{-4}\text{m}}$

Simplifying the expression:
$x=\frac{30\times {10}^{-7}}{3\times {10}^{-4}}\text{m}$

$x=10\times {10}^{-3}\text{m}$

Since ${10}^{-3}\text{m}=1\text{mm}$, we have:
$x=10\text{mm}$

Thus, the value of $x$ for the third maxima is 10 mm.