A beam of light consisting of two wavelengths $7000 \overset{0}{A}$ and $5500 \overset{0}{A}$ is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is 2.5 mm and the distance between the place of slits and the screen is 150 cm. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is $n \times 10^{- 5} m$ . The value of n is ______.

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answer is 462.

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Detailed Solution

Let $n_{1} 7000 \overset{0}{A}$ coincides with n2 maxima of $5500 \overset{0}{A}$
Therefore $n_{1} β_{1} = n_{2} β_{2}$
or $\frac{n_{1}}{n_{2}} \frac{λ_{2}}{λ_{1}} = \frac{5500}{7000} = \frac{11}{14}$
therefore 11th maxima of $7000 \overset{0}{A}$ will coincide with 14th maximum of $5500 \overset{0}{A}$
To find the least distance of this
$y = n_{1} β_{1}$
or $y = \frac{n_{1} λ_{1} D}{d} = \frac{11 \times 7000 \times 10^{- 10} \times 150 \times 10^{- 2}}{2.5 \times 10^{- 3}} = \frac{11 \times 7 \times 5}{2.5} \times 10^{5} m$
or $y = 462 \times 10^{- 5} m$
therefore n = 462

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