Yellow light of wavelength `6000Å` produces fringes of width `0.8 mm` in YDSE. What will be the fringe width if the light source is replaced by another monochromatic source of wavelength `7500 Å` and the separation between the slits is doubled ?

Explanation

Given, wavelenght of yellow light , `lambda_1 = 6000 Å`
Fringe width , `beta_1 = 0.8 mm`
separation between the slits = d
Wavelength of monochromatic source, `lambda_2 = 7500 Å`
Fringe width, `beta_2 = ?`
Separation between the slits = 2d
For the first case,
`beta_1= (D lambda_1)/(d)` (where D is the distance between screen and slit)
For the second case,
`beta_2 = (D lambda_2)/(2d)`
`:. (beta_2)/(beta_1) = (D lambda_2//2d)/(D lambda_1//d) = (lambda_2)/(2lambda_1)`
`implies beta_2 = (lambda_2)/(2lambda_1) xx beta_1`
`= (7500 Å)/(2 xx 6000 Å) xx 0.8 mm = 0.5 mm`.