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A screen is placed 50 cm from a single slit, which is illuminated with 6000 dotA light. If distance between the first and third minima in the diffraction pattern is 3.00mm, 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 lambda`. Where d is the aperture size and for small `theta : sin theta = theta =(y"/"D)` `therefore d(y/D)= n lambda, i.e., y= (D)/(d)(n lambda)` So that, `y_(3)-y_(1)= (D)/(d)(3lambda- lambda)= (D)/(d)(2lambda)` and hence, `d= (0.50xx(2xx6xx10^(-7)))/(3xx 10^(-3))= 2xx 10^(-4)m = 0.2 mm`. |
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