Question

In a certain double slit experimental arrangement interference fringes of width $1.0$ $mm$ each are observed when light of wavelength $5000$ $\mathring{A}$ is used. Keeping the set up unaltered, if the source is replaced by another source of wavelength $6000$ $\mathring{A}$, the fringe width will be

• A. $0.5$ $mm$
• B. $1.0$ $mm$
• C. $1.2$ $mm$
• D. $1.5$ $mm$

Answer 1

The correct option is C $1.2$ $mm$

$\beta =\frac{D\lambda }{d}$

${\beta }_1=\frac{D{\lambda }_1}{d}$

${\beta }_2=\frac{D{\lambda }_2}{d}$

${\beta }_2=\frac{{\lambda }_2}{{\lambda }_1}\times {\beta }_1$

$⟹{\beta }_2=\frac{6000}{5000}\times 1$

$⟹{\beta }_2=1.2$ $mm$