Rankers Physics

Young's Double Slit Experiment: Practice Problem & Solution

In Young's experiment, two coherent sources are placed 0.90 mm apart and fringe are observed one metre away. If it produces second dark fringe at a distance of 1 mm from central fringe, the wavelength of monochromatic light is used would be: (1991)
$60 \times 10^{-4} cm$
$10 \times 10^{-4} cm$
$10 \times 10^{-5} cm$
$6 \times 10^{-5} cm$

Solution Explained:

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 dark fringe is $y_n = \frac{(2n-1)\lambda D}{2d}$. For 2nd dark fringe ($n=2$), $y = \frac{3\lambda D}{2d}$. Given $y = 1 mm = 10^{-3} m$, so $10^{-3} = \frac{3 \times \lambda \times 1}{2 \times 0.90 \times 10^{-3}}$. Solving gives $\lambda = 0.6 \times 10^{-6} m = 6 \times 10^{-5} cm$.

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