InterviewSolution
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InterviewSolution
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In a single slit diffraction experiment first minimum for,lamda_1 = 660nmcoincides with first maxima for wavelength lamda_2. Calculate lamda_2 . |
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Answer» SOLUTION :Position of minima in DIFFRACTION pattern is given by, ` d sin THETA =n lamda` For first minima of `lamda_1` , we have `d sin theta_1 = (1) lamda_1 " or " sin theta_1 =(lamda_1)/(d)`.....(i) The first maxima approximately lies between first and second minima. For wavelength `lamda_2`its position will be, `d sin theta_2 = 3/2 lamda_2 therefore sin theta_2 = (3 lamda_2)/(2d)`...(II) The two will coincide if , `theta_1 = theta_2 " or " sin theta_1 = sin theta_2 therefore (lamda_1)/(d) =(3lamda_2)/(2d)` `lamda_2 = 2/3 lamda_1= 2/3 x 660 NM = 440 nm` |
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