Young's Double Slit Experiment: Practice Problem & Solution
In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is K, ($\lambda$ being the wavelength of light used). The intensity at a point where the path difference is $\lambda/4$, will be: (2014)
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:
Phase difference $\phi = \frac{2\pi}{\lambda} \times \Delta x$. When $\Delta x = \lambda$, $\phi = 2\pi$, intensity $I = I_{max}\cos^2(\pi) = I_{max} = K$. When $\Delta x = \lambda/4$, $\phi = \pi/2$. The intensity is $I = I_{max}\cos^2(\phi/2) = K\cos^2(\pi/4) = K(1/\sqrt{2})^2 = K/2$.
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