In a Young's double slit experiment, let A and B be the two slits. A thin film of thickness t and refractive index $μ$ is placed in front of A. Let $β$ = fringe width the central maximum will shift
Answer & explanation
Correct answer: option 3
Let d = distance between the slits, λ = wavelength of light,
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 = $Δ = x\frac{d}{D}$
Path difference introduced due to film = $t(μ - 1)$.
For central maximum at P, $x\frac{d}{D} = t(μ -1)$ or $x = t(μ - 1) \frac{D}{d}$
Now, $β = \frac{λD}{d}$ or $\frac{D}{d} = \frac{β}{λ}$
$∴x= t(μ - 1)\frac{β}{λ}$ towards A.