Question

In a Young’s double slit experiment, an angular width of the fringe is 0.35° on a screen placed at
2 m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index $\frac{7}{5}$, is $\frac{1}{\alpha}$. The value of $\alpha$ is _________.

Answer 1

Angular fringe width
$θ=\frac{λ}{D}$
So
$\frac{θ1}{λ1}=\frac{θ2}{λ2}$
$θ_2=\frac{0.35^∘}{450 nm}×\frac{450 nm}{715}=0.25^∘=\frac{1}{4}$
So, the value of α = 4