Question

Interference fringes were produced in Young's double slit experiment using light of wavelength 5000Å. When a film of material 2.5×10^{-3} cm thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 20 fringe widths. The refractive index of the material of the film is

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Answer 1

Fringe width, β = λD/d ... [i]

Where D is the distance between the screen and slit and d is the distance between two slits.

When a film of thickness t and refractive index μ is placed over one of the slit, the fringe pattern is shifted by distance S and is given by

$\sf:\implies\:S=\dfrac{(\mu-1)tD}{d}\dots [ii]$

Given : S = 20β ... [iii]

From equations [i], [ii] and [iii] we get,

$\sf:\implies\:(\mu-1)t=20\lambda$

$\sf:\implies\:(\mu-1)=\dfrac{20\lambda}{t}$

$\sf:\implies\:(\mu-1)=\dfrac{20\times5000\times10^{-8}\:cm}{2.5\times10^{-3}\:cm}$

$\sf:\implies\:\mu-1=0.4$

$\sf:\implies\:\mu=0.4+1$

$:\implies\:\underline{\boxed{\bf{\orange{\mu=1.4}}}}$

Answer 2

Answer:

ringe width, β = λD/d ... [i]

Where D is the distance between the screen and slit and d is the distance between two slits.

When a film of thickness t and refractive index μ is placed over one of the slit, the fringe pattern is shifted by distance S and is given by

\sf:\implies\:S=\dfrac{(\mu-1)tD}{d}\dots [ii]:⟹S=

d

(μ−1)tD

…[ii]

Given : S = 20β ... [iii]

From equations [i], [ii] and [iii] we get,

\sf:\implies\:(\mu-1)t=20\lambda:⟹(μ−1)t=20λ

\sf:\implies\:(\mu-1)=\dfrac{20\lambda}{t}:⟹(μ−1)=

t

20λ

\sf:\implies\:(\mu-1)=\dfrac{20\times5000\times10^{-8}\:cm}{2.5\times10^{-3}\:cm}:⟹(μ−1)=

2.5×10

−3

cm

20×5000×10

−8

cm

\sf:\implies\:\mu-1=0.4:⟹μ−1=0.4

\sf:\implies\:\mu=0.4+1:⟹μ=0.4+1

:\implies\:\underline{\boxed{\bf{\orange{\mu=1.4}}}}:⟹

μ=1.4

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