Young's Double Slit Experiment: Practice Problem & Solution
In Young's double slit experiment carried out with light of wavelength ($\lambda$) = $5000 \AA$, the distance between the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum (taking the central maximum as zeroth maximum) will be at x equal to (1992)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Young's Double Slit Experiment. Understanding the underlying formula is key to arriving at the correct answer below:
Position of nth maximum is $x_n = \frac{n\lambda D}{d}$. For $n=3$, $x_3 = \frac{3 \times 5000 \times 10^{-10} \times 2}{0.2 \times 10^{-3}} = \frac{3 \times 10^{-6}}{0.2 \times 10^{-3}} = 15 \times 10^{-3} m = 1.5 cm$.
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