Question

Yellow light of wavelength $6000 \mathring{A}$ produces fringes of width $0.8\, mm$ in Youngs double slit experiment. If the source is replaced by another monochromatic source of wavelength $7500 \mathring{A}$ 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

Answer: (B)

Fringe width in first case, $\beta_{1} = \frac{D\lambda_{1}}{d}...\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 \mathring{A}}{6000 \mathring{A}} \times 0.8 mm $
$= 0.5 mm $