Q.In a biprism experiment, the fringes are observed in the focal plane of the eyepiece at a distance of 1.2 m from the slits. The distance between the central bright band and the 20th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eyepiece, 90 cm from the eyepiece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used.

[!TLDR] Using the lens conjugate-foci method, the true virtual-source separation is $d=d'\times(u/v)=0.9\text{cm}\times(30/90)=0.3$ cm, and $\lambda=W\times d/D$ with $W=0 …

This biprism setup uses the classic lens-displacement (conjugate foci) method to determine the TRUE (small) separation d of the two virtual sources, which is normally too small to measure directly. First, the fringe width is found from the directly observed spacing: the distance from the central bright band to the 20th bright band is 0.4 cm, spanning 20 fringe widths, so $\beta=0.4\text{ cm}/20=0.02\text{ cm}=2\times10^{-4}$ m. …

Get the fringe width from the given 20-fringe span, get the true (small) virtual-source separation d from the magnified image separation using the lens conjugate-foci relation $d=d'\tim …

Using the magnified image separation $d'=0.9$ cm directly as d in the fringe-width formula, instead of first converting it back to the true (unmagnif …