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