Q. In Young’s double slit experiment, the 8th maximum with wavelength λ_{1} is at a distance d_{1} from the central maximum and the 6th maximum with wavelength λ_{2} is at a distance d_{2} from central maximum. Then, d_{1}/d_{2} 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: d

_{1}= 8λ_{1}D/d , d_{2}= 6λ_{2}D/dOn dividing , d_{1}/d_{2} = (4/3)(λ_{1}/ λ_{2})