In a double slit experiment, when a thin film of thickness t having refractive index μ is introuduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value

Correct answer: D. (λ)/(( μ - 1 )) Shift = (μ - 1)t(D)/(d) Fringe width ( β ) = ( )/(d) ( μ - 1 ) t(D)/(d) = λ(D)/(d) t = (λ)/(μ - 1)

JEE Main 201912 AprilMorning Shift

JEE Main 2019 · 12 April · Morning Shift · Physics · Wave Optics

In a double slit experiment, when a thin film of thickness t having...

In a double slit experiment, when a thin film of thickness

t

having refractive index

\mu

is introuduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of

t

( \lambda

is the wavelength of the light used):

D. $\dfrac{\lambda}{( \mu - 1 )}$

Shift

= \left(\mu - 1\right)t\dfrac{D}{d} <br/>Fringe width

( \beta ) = \dfrac{\lambdaD}{d}

( \mu - 1 ) t\dfrac{D}{d} = \lambda\dfrac{D}{d}

t = \dfrac{\lambda}{\mu - 1}