Question

In a biprism experiment, light of wavelength $5200\mathring{A}$ is used to get an interference pattern on the screen. The fringe width changes by $1.3mm$ when the screen is moved towards biprism by $50cm$. Find the distance between two virtual images of the slit.

Answer 1

$\lambda =5200A^o=52\times {10}^{-8}metres$

$\Delta X=1.3mm=1.3\times {10}^{-3}m$

$\Delta D=50cm=0.5m$

$d=?$

$X_1=\frac{\lambda D_1}{d}$ and $X_2=\frac{\lambda D_2}{d}$

$\Delta X=X_1-X_2=\frac{\lambda }{d}(D_1-D_2=\Delta D)$

$\therefore d=\frac{\lambda \times \Delta D}{\Delta X}=$

$\therefore d=\frac{52\times {10}^{-8}\times 0.5}{1.3\times {10}^{-3}}=20\times {10}^{-5}=0.2\times {10}^{-3}metres=\frac{0.2mm}{}$