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In Young's double slit experiment the sl...


In Young's double slit experiment the slits, `S_(1) & S_(2)` are illuminated by a parallel beam of light of wavelength 4000 Å from the medium of refractive index `n_(1)=1.2` A thin film of thickness 1.2 `mum` and refractive index n=1.5 is placed infrom of `S_(1)` perpedicular to path of light. the refractive index of medium between plane of slits & screen is `n_(2)=1.4` if the light coming from the film and `S_(1)&S_(2)` have equal intensities I then intensity at geometrical centre of the screen O is

A

0

B

2I

C

4I

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
Path difference at O: `(mu_(rel)-1)t=((n)/(n_(2))-1)t`
`therefore `phase difference at O: `t((n)/(n_(2))-1)(pi)/(lamda_(2))` where `n_(1)lamda_(1)-n_(2)lamda_(2)`
`implies` phase difference `=(pi)/(2)implies"Resultant intensity "=2I`
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Knowledge Check

  • A light wave enters from air into a medium of refractive index 1.5 . The speed of light in the medium will be

    A
    `2 xx 10 ^(8) m// s`
    B
    ` 4 . 5 xx 10 ^(8) m// s`
    C
    ` 3 xx 10 ^(8) m// s`
    D
    none of these

  • A light wave enters from air into a medium of refractive index 1.5. The speed of light in the medium will be

    A
    `2 xx 10^(8)m//s`
    B
    `4.5 xx 10^(8) m//s`
    C
    `9 xx 10^(8) m//s`
    D
    `(330//1.5) xx 10^(8) m//s`

  • A double slit experiment is performed with light of wavelength 500 nm A thin film of thickness 2 mu m and refractive index 1.5 is introduced in the path of the upper beam the location of the central maximum will

    A
    remain unshifted
    B
    shift downward by nearly two fringes
    C
    shift upward by nearly two fringes
    D
    shift downward by its fringes

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