In Young's double slit experiment, the slit are 2 mm apart and are illuminated by photons of two wavelengths `lambda = 12000 Å and lambda_(2)=10000 Å`. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

Options

A. 4 mm
B. 3 m
C. 8 mm
D. 6 mm

Answer

D

Explanation

Suppose n, th bright fringe of wavelength `lambda_(1) and n_(2)th` bright fringe of wavelength `lambda_(2)` are superposing to each other.
`:.n barX_(1)=n_(2)barX_(2)`
`:.(n_(1)lambda_(1)D)/(d)=(n_(2)lambda_(2)D)/(d)`
`:. n_(1) lambda_(1)=n_(2)lambda_(2)`
`:. (n_(1))/(n_(2))=(lambda_(2))/(lambda_(1))=(10000Å)/(12000Å)=(5)/(6)`
`:.n_(1)=5 th and n_(2)=6th` fringes are superposing Now distance of central fringe from mid point
`x=(n_(1) lambda_(1)D)/(d)`
`=5xx12000xx10^(-4)xx(2)/(2xx10^(-3))`
`=6000xx10^(-7)`
`=6xx10^(-3)m=6 mm`
This distance can be determine by `x=(n_(2) lambda_(2)D)/(d)` also.