BITSAT 2015 Paper

Given: fringe with b = 0.03 cm, D = 1 m = 100 cm Distance between images of the source = 0.8 cm. Image distance v = 80 cm Object distance = u Using mirror formula $\frac{1}{v} + \frac{1}{u}$ = $\frac{1}{f}$ ⇒ $\frac{1}{60} + \frac{1}{u}$ = $\frac{1}{16}$ ⇒ u = 20 cm Magnification, m = $\frac{v}{u}$ = $\frac{80}{20}$ = 4 Magnification = $\frac{\text{distances between images of slits}}{\text{distance between slits}}$ = $\frac{0.8}{d}$ = 4 ⇒ d = 0.2 cm Fringe width β = $\frac{D\lambda}{d}$ or , β = $\frac{100\lambda}{0.2}$ = 0.03 × $10^{-2}$ Therefore, wavelength of light used λ = 6000 Å