In Young’s double slit experiment, the fringe width is ß. If the entire arrangement is placed in a liquid of refractive index n, the
fringe width becomes :
(A) nß (B) +1
ß
n (C) -1
ß
n (D) n
ß

Question

In Young’s double slit experiment, the fringe width is ß. If the entire arrangement is placed in a liquid of refractive index n, the fringe width becomes : (A) nß (B) +1 ß n (C) -1 ß n (D) n ß

Arun · Answer

If the screen distance id D, distance between the slits is d and the wave length is λ
then the fringe width is given by
b = λD/d
When the arrangement is immerse in a liquid of refractive index u then the new wavelength is 1/u times the original wavelength
If the new wavelength is λ' then
λ' = λ/n
Therefore, the new fringe width will be
b' = λ'D/d

or, b' = b/u
Therefore the fringe width will decrease by a factor of u.

Vikas TU · Answer

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In Young’s double slit experiment, the fringe width is ß. If the entire arrangement is placed in a liquid of refractive index n, the fringe width becomes : (A) nß (B) +1 ß n (C) -1 ß n (D) n ß

In Young’s double slit experiment, the fringe width is ß. If the entire arrangement is placed in a liquid of refractive index n, the
fringe width becomes :
(A) nß (B) +1
ß
n (C) -1
ß
n (D) n
ß

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In Young’s double slit experiment, the fringe width is ß. If the entire arrangement is placed in a liquid of refractive index n, the fringe width becomes : (A) nß (B) +1 ß n (C) -1 ß n (D) n ß

In Young’s double slit experiment, the fringe width is ß. If the entire arrangement is placed in a liquid of refractive index n, the fringe width becomes : (A) nß (B) +1 ß n (C) -1 ß n (D) n ß

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In Young’s double slit experiment, the fringe width is ß. If the entire arrangement is placed in a liquid of refractive index n, the
fringe width becomes :
(A) nß (B) +1
ß
n (C) -1
ß
n (D) n
ß