Q.The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000 Å is 0.32 mm by how much will the distance change if light of wavelength 4800 Å is used?

[!TLDR] Fringe width scales directly with wavelength ($W\propto\lambda$, everything else fixed), so $W_2=W_1\times(\lambda_2/\lambda_1)=0.32\times(4800/6000)=0.256$ mm, a decrease …

In a biprism (or any double-slit-equivalent) setup, the fringe width $W=\lambda D/d$ is directly PROPORTIONAL to the wavelength $\lambda$, with D and d both fixed properties of the unchanged physical apparatus. So for two different wavelengths using the SAME apparatus, $\dfrac{W_2}{W_1}=\dfrac{\lambda_2}{\lambda_1}$. …

Use the direct proportionality between fringe width and wavelength ($W\propto\lambda$, with D and d unchanged) to scale the given fringe width by the …

Reporting the new fringe width (0.256 mm) as the final answer, rather than correctly reporting the CHANGE (0.064 mm) that the question actually asks for; or forgetting the change is a de …