A beam of light consisting of wavelengths 6000 `Å` and 4500 `Å` is used in a YDSE with D = 1 m and d = 1 mm. Find the least distance from the central maxima, where bright fringes due to the two wavelengths coincide.

Options

A. 0.9 mm
B. 1.2 mm
C. 1.8 mm
D. 1.4 mm

Answer

C

Explanation

`beta _(1) = ( lamda _(1)D)/( d) =( 6000 xx 10^(-10) xx1)/( 10 -3)= 0.6mm`
` beta 2 =( lamda _(2) D) /( d) = 0.45 mm`
Let `n_(1) ` th maxima of `lamda _(1) and n_(2) `th maxima of `lamda _(2) ` coincide at a position y.
then , `y= n_(1) beta _(1) = n_(2) beta _(2) = LCM` of `beta _(1) beta _(2) `
`implies y= LCM` of 0.6 mm and 0.45 mm
`y= 1.8 mm`
At this point 3rd maxima for 6000 Å & 4th maxima for 4500 Å coincide