In a YDSE perfromed with light of wavelength `600 Å`, the screen is placed 1 m from the slits. Fringes formed on the screen are observed by a student sitting close to the slits. The student's eye can distinguish two neighboring fringes. If they subtend an angle more then 1 minute of ace, then
In order to have the clear visibility of the fringe, the maximum distance that can be maintained between the slits is

Options

A. `8.74 xx 10^(-4) mm`
B. `6.74 xx 10^(-4) mm`
C. `5 xx 10^(-4) mm`
D. `8.74 xx 10^(-7) mm`

Answer

c

Explanation

The angular fringe width, i.e., the angle subtended by a fringe at the center of slits is given by
` beta_(theta) = (beta)/(D) = (lambda)/(d)`
According to the given conditions,
`beta_(theta) = (lambda)/(d) ge (pi)/(180 xx 60)`
`d le (6 xx 10^(-7) xx 180 xx 60)/(pi)`
`d le 2.06 xx 10^(-3) m = 2.06 mm`
Position of 3rd bright fringe,
`y_(3) = (3 lambda D)/(d) = (3 xx 6 xx 10^(-7) xx 1)/(2.06 xx 10^(-3)) = 8.74 xx 10^(-4) m`
Position of 5th dark fringe,
`y_(5) = ((n - (1)/(2)) lambda D)/(d) = (9 lambda D)/(2 d)`
`= (9 xx 6 xx 10^(-7) xx 1)/(2 xx 2.06 xx 10^(-3)) = 13.1 xx 10^(-4) m = 1.31 mm`