A lens inserted between the Fresnel biprism and eye-piece which are 1 m apart gives two images of the slit in two positions. In first case, the images of the slit are `9 xx 10^(-3)m` and in the second position the images are `4 xx 10^(-3)m` apart. The distance between the interference fringes is found to be `1 xx 10^(-M)` m when sodium light of wavelength `6000 xx 10^(-10)` m is used. What is the value of M? (Take `pi = 3.14`)

Answer

`04.00`

Explanation

When the distance between the screen and the object is greater than four times the focal length of a convex lens, there are two positions of the lens where it forms a real image of the object on the screen. If `d_1 and d_2` are the sizes of the images in the two positions then d (size of the object `= sqrt(d_1d_2)`
`:.` d(distance between virtual sources)
`= sqrt(9 xx 10^(-3) xx 4 xx 10^(-3))`
But, `beta = (lamdaD)/d `
`:. beta = (6000 xx10^(-10)xx1)/(sqrt(4xx9 xx10^(-6)))= (6000 xx10^(-10))/(6 xx10^(-3))=1 xx 10^(-4)`
`:. M = 4`