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In Young's double slit experiment, the slit are 2 mm apart and are illuminated by photons of two wavelengths lambda = 12000 Å andlambda_(2)=10000 Å. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other? |
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Answer» 4 mm `:.n barX_(1)=n_(2)barX_(2)` `:.(n_(1)lambda_(1)D)/(d)=(n_(2)lambda_(2)D)/(d)` `:. n_(1) lambda_(1)=n_(2)lambda_(2)` `:. (n_(1))/(n_(2))=(lambda_(2))/(lambda_(1))=(10000Å)/(12000Å)=(5)/(6)` `:.n_(1)=5 th and n_(2)=6th` fringes are superposing Now distance of central fringe from mid point `x=(n_(1) lambda_(1)D)/(d)` `=5xx12000xx10^(-4)xx(2)/(2xx10^(-3))` `=6000xx10^(-7)` `=6xx10^(-3)m=6 mm` This distance can be determine by `x=(n_(2) lambda_(2)D)/(d)` also. |
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