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Light of wavelength 6000 Å is reflected ...

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

A

infinity

B

`200 Å`

C

`2000 Å`

D

`1000 Å`

Text Solution

Verified by Experts

The correct Answer is:
C
(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`
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Knowledge Check

  • A parallel beam of sodium light of wavelength 6000 Å is incident on a thin glass plate of mu=1.5 , such that the angle of refraction in the plate is 60^(@) . The smallest thickness of the plate which will make it appear dark by reflected light is

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    `40 Å`
    B
    `4 Å`
    C
    `400 Å`
    D
    `4000 Å`

  • The central fringe of the interference pattern produced by the light of wavelength 6000 Å is found to shift to the position of 4th dark fringe after a glass sheet of refractive index 1.5 is introduced. The thickness of glass sheet would be

    A
    `4.8 mu m`
    B
    `4.2 mu m`
    C
    `5.4 mu m `
    D
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  • In double slit experiment fringes are obtained using light of wavelength 4800 Å One slit is covered with a thin glass film of refractive index. 1.4 and another slit is covered by a film of same thickness but refractive index 1.7 . By doing so, the central fringe is shifted to fifth bright fringe in the original pattern. The thickness of glass film is

    A
    `2xx10^(-3)mm`
    B
    `4xx10^(-3)mm`
    C
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    D
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