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In a single slit diffraction experiment first minimum for lambda_(1)=66nm coincides with first maxima for wavelength lambda_(2). Calculate lambda_(2). |
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Answer» Solution :Position of minima in diffraction PATTERN is give by, `d sin theta = n lambda`. For first minima of `lambda_(1)`, we have `d sin theta_(1)= (1)lambda_(1)" or "sin theta_(1)= (lambda_1)/(d)"""……….."(i)` The first manima approximately LIES between first ans second minima. For wavelength `lambda_(2)` its position will be, `d sin theta_(2)= (3)/(2)lambda_(2)""therefore sin theta_(2)= (3 lambda_2)/(2d)"""..........."(ii)` The TWO will coincide if, `theta_(1)=theta_(2)" or "sin theta_(1)= sin theta_(2)` `therefore (lambda_1)/(d)= (3lambda_(2))/(2d)" or "lambda_(2)= (2)/(3)lambda_(1)= (2)/(3)xx 660 nm= 440nm`. |
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