Young's Double Slit Experiment: Practice Problem & Solution
In Young's double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths $\lambda_1 = 12000 \AA$ and $\lambda_2 = 10000 \AA$. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other? (2013)
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:
For coincidence, $n_1\lambda_1 = n_2\lambda_2 \Rightarrow n_1(12000) = n_2(10000) \Rightarrow \frac{n_1}{n_2} = \frac{5}{6}$. Minimum values are $n_1=5$, $n_2=6$. The minimum distance $y = \frac{n_1\lambda_1 D}{d} = \frac{5 \times 12000 \times 10^{-10} \times 2}{2 \times 10^{-3}} = 6 \times 10^{-3} m = 6 mm$.
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