Question

Yellow light of wavelength $6000\mathring{A}$ produces fringes of width $0.8mm$ 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. $0.1mm$
• B. $0.5mm$
• C. $4.3mm$
• D. $1mm$

Answer 1

Answer: (B)

Fringe width in first case, ${\beta }_1=\frac{D{\lambda }_1}{d}...\left(i\right)$
Fringe width in second case, ${\beta }_2=\frac{D{\lambda }_2}{2d}...\left(ii\right)$
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
${\beta }_2=\frac{1}{2}\cdot \frac{{\lambda }_2}{{\lambda }_1}\cdot {\beta }_1$
$\therefore {\beta }_2=\frac{1}{2}\times \frac{7500\mathring{A}}{6000\mathring{A}}\times 0.8mm$
$=0.5mm$