Question
The maximum intensity in a Young's double slit experiment is $I_{0}$. Distance between the slits (d) is $5\lambda$, where $\lambda$ is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at $D = 10\text{ }d$ is $\_\_\_\_$
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Solution
Sol. [Key Concept: The path difference opposite one slit gives $\Delta\varphi$; then $I = I_{0}\cos^{2}\frac{\Delta\varphi}{2}$.]
At $Py = \frac{d}{2}$
$${\Delta Q = \frac{2\pi}{\lambda}\frac{yd}{D} = \frac{2\pi}{\lambda}\frac{\text{ }d}{2}\frac{\text{ }d}{D} = \frac{\pi d^{2}}{\lambda D} }{\Delta Q = \frac{\pi \times 25\lambda^{2}}{\lambda \times 10\text{ }d} = \frac{5\pi\lambda}{2 \times 5\lambda} = \frac{\pi}{2} }{I = I_{0}\cos^{2}\left( \frac{\Delta Q}{2} \right) = I_{0}\cos^{2}\left( \frac{\pi}{2 \times 2} \right) = \frac{I_{0}}{2} }$$
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