Light of wavelength `6000 Å` is reflected at nearly normal incidence from a soap films of refractive index 1.4 The least thickness of the fringe then will appear black is

Options

A. infinity
B. `200 Å`
C. `2000 Å`
D. `1000 Å`

Answer

C

Explanation

(c ) Black fringe means destructive interference. For destructive interference, we have
`2 mu t cos r = n lambda`
Where `mu` is refractive index, `lambda` is wavelength and `t` is thichness
`implies t = (n lambda)/(2 mu cos r)`
For minima, we have `cos r = 1, n = 1`
`implies t = (lambda)/(2 mu) = (6000)/(2 xx 1.4) = 2142 Å`
`~~ 2000 Å`