Question

Yellow light of wavelength $6000 ?$ produces fringes of width $0.8\, mm$ in Youngs double slit experiment. If the source is replaced by another monochromatic source of wavelength $7500 ?$ and the separation between the slits is doubled then the fringe width becomes

• A. $0.1\,mm $
• B. $0.5\,mm $
• C. $4.3\,mm $
• D. $1\,mm $

Answer 1

The Correct Option is B

Fringe width in first case, $\beta_{1} = \frac{D\lambda_{1}}{d}\quad...\left(i\right) $ Fringe width in second case, $\beta_{2}=\frac{D\lambda_{2}}{2d} \quad...\left(ii\right) $ Divide equation $\left(ii\right)$ by $\left(i\right)$, $ \therefore \frac{\beta_{2}}{\beta_{1}}= \frac{D\lambda_{2}/ 2d}{D\lambda_{1} /d} = \frac{1}{2} \cdot\frac{\lambda_{2}}{\lambda_{1}} $ or $\beta_{2} = \frac{1}{2}\cdot\frac{\lambda_{2}}{\lambda_{1}} \cdot\beta_{1} $ $\therefore \beta_{2} = \frac{1}{2}\times\frac{ 7500 ?}{6000 ?} \times 0.8 mm $ $= 0.5 mm $