In Young's experiment, interference pattern is obtained on a screen at a distance of `1 m` from slits separated by `0.05 cm` and illuminated by sodium light of wavelength `5893 Å`. Calculate distance between `4th` bright fringe on one side and 3rd bright fringe on other side of central bright fringe.

Answer

`8.25 xx 10^(-3) m`

Explanation

Here, `D = 1 m, d = 0.05 cm = 5 xx 10^(-4)m` ,brgt `lambda = 5893 Å = 5893 xx 10^(-10)m`
Distance between 4th bright fringe on one side and 3rd bright fringe on other side of central max.
is equal to width of `7` fringes `= 7(lambda D)/(d)`
`= (7 xx 5893 xx 10^(-10) xx 1)/(5 xx 10^(-4))m = 8.25 xx 10^(-3)m`