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 ?

Answer

`0.5 mm`

Explanation

Here, `lambda_(1) = 6000Å, beta_(1) = 0.8 mm`,
`lambad_(2) = 7500 Å, d_(2) d_(1), beta_(2) = ?`
`beta_(1) = (lambda_(1)D)/(d_(1)) and beta_(2) = (lambda_(2)D)/(d_(2))`
`(beta_(2))/(beta_(1)) = (lambda_(2))/(lambda_(2)) xx (d_(1))/(lambda_(1))`
`beta_(2) = beta_(1)(lambda_(2))/(lambda_(1)) xx (d_(1))/(d_(2)) = 0.8 xx (7500)/(6800) xx (1)/(2) = 0.5 mm`