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A screen is placed 50 cm from a single slit, which is illuminated with 6000 Å light. If distance between the first and third minima in the diffraction pattern is 3.00 mm, what is the width of the slit ? |
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Answer» Solution :In case of diffraction at single slit, the position of minima is given by `d SIN theta= n lamda`. where d is the aperture size and for small `theta : sin theta = theta =(y//D)` `THEREFORE d(y/D) = n lamda, i.e., y = D/d (n lamda)` so that `y_3 - y_1 = D/d (3lamda -lamda) = D/d (2 lamda)` and hence ` d = (0.50xx(2 xx 6 xx 10^(-7)) )/(3 xx 10^(-3)) = 2 xx 10^(-4) m = 0.2 mm` |
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