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Q.

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

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a

$\frac{3}{4} (\frac{λ_{1}}{λ_{2}})$

b

$\frac{4}{3} (\frac{λ_{2}}{λ_{1}})$

c

$\frac{4}{3} (\frac{λ_{1}}{λ_{2}})$

d

$\frac{3}{4} (\frac{λ_{2}}{λ_{1}})$

answer is B.

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Detailed Solution

Position of n^th maxima from central maxima is Given by $x_{n} = \frac{n λ D}{d}$
For 8^th maxima $x_{6} = \frac{8 λ_{1} D}{d_{1}}$ and 6^th for maxima $x_{6} = \frac{6 λ_{2} D}{d_{2}}$
Now, $x_{6} = x_{8}; \Rightarrow \frac{d_{1}}{d_{2}} = \frac{n_{1} λ_{1}}{n_{2} λ_{2}} = \frac{4}{3} (\frac{λ_{1}}{λ_{2}})$

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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