A beam of light consisting of two wavelengths, 650 nm and 520 nm is used to obtain interference fringes in Young's double-slit experiment. What is the least distance (in m) from a central maximum where the bright fringes due to both the wavelengths coincide ? The distance between the slits is 3 mm and the distance between the plane of the slits and the screen is 150 cm.

Options

A. `0.16 cm`
B. `0.32 cm`
C. `0.48 cm`
D. `1.92 cm`

Answer

A

Explanation

To find the point of coincidence of bright fringes we can equate the distance of bright fringes from the central maxima made by both the wavelengths of light
Given, distance between the screen and slit
`D=120cm =120xx10^(-2)m`
Slit width `d=2mm=2xx10^(-3)m`
`lambda_(1)=6500 Å and lambda_(2)=5200 Å`
Let nth fringe due to `lambda_(2)=5200 Å` coincide with `(n-1)` th bright fringe due to `lambda_(1)=6500 Å`
`:. (Dnlambda_(2))/d=(D(n-1)lambda_(1))/dor nxx5200=(n-1)6500`
or `4n=5n-5 rArr n=5`
So, the least distance required `y_(n)=(nlambda_(2)D)/d`
`=(5xx5200xx10^(-10)xx120xx10^(-2))/(2xx20^(-3))=0.16xx10^(-2)m=0.16cm`