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  • In a Young's double slit experiment, let A and B be the two slits. A thin film of thickness t and refractive index i

In a Young's double slit experiment, let A and B be the two slits. A thin film of thickness t and refractive index \muis placed in front of A. Let \beta =  fringe width the central maximum will shift 

Option: 1

towards A


Option: 2

towards B


Option: 3

by \, t(\mu-1) \frac{\beta}{\lambda}


Option: 4

by \: \mu \mathrm{t} \frac{\beta}{\lambda}


Answers (1)

best_answer

 Let \mathrm{d=} distance between the slits, \mathrm{\lambda=} wavelength of light,
\mathrm{D=} distance from the slits to the screen.

For a point P on the screen at a distance x from the centre of the screen, path difference \mathrm{=\Delta=\mathrm{x}\frac{\mathrm{d}}{\mathrm{D}}}
Path difference introduced due to film \mathrm{=\mathrm{t}(\mu-1)}.
For central maximum at \mathrm{P, x \frac{d}{D}=t(\mu-1) \quad\, or \, x=t(\mu-1) \frac{D}{d}}
\mathrm{Now, \beta=\frac{\lambda D}{d} \quad or \quad \frac{D}{d}=\frac{\beta}{\lambda}}
\mathrm{\therefore \quad \mathrm{x}=\mathrm{t}(\mu-1) \frac{\beta}{\lambda} \, towards \, A.}

 

Posted by

Anam Khan

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