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In Young' s double-slit experiment, separation between the two slits is d and separation between plane of the slits and the screen is D. A small hole is made on the screen at a point directly in front of one of the slit. Some wavelengths are missing from the light coming out from hole on the other side. These wavelengths are |
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Answer» `(d^2)/D` `Deltax = (yd)/D` For the point directly in front of ONE of the slits `y = d//2`, hence on substituting we get path difference as follows: `Delta x = (yd)/D = ((d/2)d)/D` `implies Delta x = (d^2)/(2D)` There will be destructive interference of those wavelengths which are missing, at this point and condition for destructive interference can be written as follows: `Delta x = (2n + 1)lambda/2` Here, `lambda` is wavelength od light `implies (d^2)/(2D) = (2n + 1) lambda/2` `implies lambda = (d^2)/((2n + 1)D)` We can substitute `n = 0, 1, 2, 3,............` `implies lambda = (d^2)/(D), (d^2)/(3D) , (d^2)/(5D),.....` |
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