In a double slit experiment, the distance between slits is increased 10 times whereas their distance from screen is halved, then what is the fringe width?

Options

A. It remains same
B. Becomes 1/20
C. Becomes 1/10
D. Becomes 1/90

Answer

C

Explanation

Let A be wavelength of monochromatic light, d the distance between coherent sources, and D the distance between screen and source, then fringe width is

`beta=(Dlambda)/(d)`
Given, `d_(1)=d, D_(1)=D, d_(2)=10d, D_(2)=(D)/(2)`
`therefore beta_(2)=((D)/(2)lambda)/(10d)=(Dlambda)/(20d)`
`Rightarrow beta_(2)=(W)/(20)`