(a) In Young's double-slit experiment, derive the condition for
(i) constructive interference and (ii) destructive interference ata point on the screen.
(b) A beam of light consisting of two wavelenths, 800 nm and 600 nm is used to obtain the interference fringes in a Young's double-slit experiment on a screen placed 1.4m away. if the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.

Explanation

(b) Here `lamda_(1)=800nm,lamda_(2)=600nm,D=1.4m and d=0.28mm=2.8xx10^(-4)m`.
Let at a distance .x. from central maxima the bright fringes due to `lamda_(1) and lamda_(2)` coincide first time. For this to happen
`nlamda_(1)=(n+1)lamda_2`, where n is an integer
`implies (n+1)/(n)=(lamda_(1))/(lamda_(2))=(800nm)/(600nm)=4/3 implies n=3`.
It means that at distance x, nth i.e., 3rd maxima for wavelength `lamda_(1)` is just coinciding with (n+1)th i.e., 4th maxima for wavelength `lamda_(2)`
`therefore x=(nDlamda_(1))/(d)=(3xx1.4xx800xx10^(-9))/(2.8xx10^(-4))=1.2xx10^(-2)m or 1.2cm`.