InterviewSolution
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InterviewSolution
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A plane light wave with wavelength lambda = 0.54 mu m goes through a thin converging lens with focal length f = 50 cm and an aperture stop fixed immediately after lens, and reaches a screen placed at a distance b = 75 cm from the aperture stop. At what aperture radii has the centre of the differaction pattern on the screen the maximum illuminance ? |
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Answer» Solution :Here the focal point acts as a virtual source of LIGHT. This MEANS that we can take spherical waves converging towards `F`. Let us divide these waves into Fresnel zones just after they energy from the stop. We write `r^(2) = f^(2) - (f -h)^(2) = (b - mlambda//2)^(2) - (b - h)^(2)` Here `r` is the radius of the `m^(th)` fresnel zone and `h` is the distance to the left of the food of the PERPENDICULAR. Thus `r^(2) = 2fh = -bm lambda + 2bh` So `h = bm lambda//2(b - f)` and `r^(2) = fbm lambda//2(b - f)`. The intensity maxima are observed when an oss number of Fresnel zones are exposed by the stop. Thus `r_(k) = SQRT((kbf lambda)/(b - f))` where `k = 1, 3, 5.......` 
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