Q. In Young’s double slit experiment, the 8th maximum with wavelength λ1 is at a distance d1 from the central maximum and the 6th maximum with wavelength λ2 is at a distance d2 from central maximum. Then, d1/d2 is equal to
$\displaystyle (a) \frac{4}{3}\frac{\lambda_2}{\lambda_1} $
$\displaystyle (b) \frac{4}{3}\frac{\lambda_1}{\lambda_2} $
$\displaystyle (c) \frac{3}{4}\frac{\lambda_2}{\lambda_1} $
$\displaystyle (d) \frac{3}{4}\frac{\lambda_1}{\lambda_2} $
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Ans: (b)
Sol: d1 = 8λ1D/d , d2 = 6λ2D/d
On dividing , d1/d2 = (4/3)(λ1/ λ2)