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Physics Question on Youngs double slit experiment

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.

Answer

β=λDd=5×10−7×23×10−4=10×10−3 m\beta = \frac{\lambda D}{d} = \frac{5 \times 10^{-7} \times 2}{3 \times 10^{-4}} = 10 \times 10^{-3} \, \text{m}

For 3rd maxima:

y3=3β=10×10−3 m=10 mmy_3 = 3\beta = 10 \times 10^{-3} \, \text{m} = 10 \, \text{mm}