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State the condition under which the phenomenon of diffraction of light takes place. Derive an expression for the width of the central maximum due to diffraction of light at a single slit. A slit of width 'a' is illuminated by a monochromatic light of wavelength 700 nm at normal incidence. Calculate the value of 'a' for position of (i) first minimum at an angle of diffraction of 30. (ii) first maximum at an angle of diffraction 30. |
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Answer» Solution :`lambda = 700 nm = 700 xx 10^(-9) m` `theta = 30^@` (i) For first minima `a SIN theta = n lambda` `n = 1 , a sin 30 = lambda` `a = (lambda)/(sin 30^@) = 2lambda = 2 xx 700 xx 10^(-9)` `= 14 xx 10^(-7) m`. (ii) For first maxima, `a sin theta = (2n + 1) lambda/2` `a sin 30^@ = (3lambda)/2` `a = (3 lambda)/(2 xx sin 30^@) = 3 lambda = 3 xx 700 xx 10^(-9)` `= 21 xx 10^(-7) m`. |
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