In a Young's double slit experiment, slits are separated by `0.5 mm` and the screen is placed `150 cm` away. A beam of light consisting of two wavelengths, `650 nm` and `520 nm`, is used to obtain interference fringes on the screen. The least distance from the commom central maximum to the point where the bright fringes fue to both the wavelengths coincide is

Options

A. 1.56 mm
B. 7.8 mm
C. 9.75 mm
D. 15.6 mm

Answer

B

Explanation

Wavelength - 1, `lambda_1 = 650 nm`
Wavelength - 2 `lambda_2 = 520 nm`
Distance between the slits, `d = 0.5 mm = 5 xx 10^(-4)m`
Distance of the slits from the screen , `D = 150 cm = 1.5 m`
The central maxima for both the interference patterns will be at same postion.
Let `n_1^(th)` bright fringe corresponding to wavelength-1 coincide with `n_2^(th)` bright fringe for the wavelength - 2.
Distance of `n_1^(th)` bright fringe corresponding to wavelength- 1 from the central maxima , `x_1 = (n_1D lambda_1)/(d)`
Distance of `n_2^(th)` bright fringe corresponding to wavelength-2 from the central maxima, `x_2 = (n_2D lambda_2)/(d)`
`x_1 = x_2`
`(n_1 D lambda_1)/(d) = (n_2D lambda_2)/d`
`n_1lambda_1 = n_2lambda_2`
`(n_1)/(n_2) = (lambda_2)/(lambda_1) = (520 xx 10^(-9))/(650 xx 10^(-9)) = 4/5`
The minimum distance from the central bright fringe is given by :
`x_1 = (4 xx 1.5 xx 650 xx 10^(-9))/(5 xx 10^(-4)) = 780 xx 10^(-5) = 7.8 mm`