Question

We get $60$ fringes in the field of view of monochromatic light of wavelength $4000\stackrel{{}^{\circ }}{A}$ in Young's double-slit experiment. What will be the number of fringes obtained in the same field of view, if we use monochromatic light of wavelength $6000\stackrel{{}^{\circ }}{A}$ .

• A. $60$
• B. $90$
• C. $40$
• D. $1.5$

Answer 1

Answer: (C)

$\beta =\frac{\lambda D}{d}$
When monochromatic light of wavelength $4000A$ is used:
$60th$ fringe is obtained at $=60\beta$
$=60\frac{{\lambda }_{1}D}{d}$ ......eq 1
When, monochromatic light of wavelength $6000A$ is used,
${\beta }_2=\frac{{\lambda }_{2}D}{d}$
$nth$ fringe is at distance = $n(\frac{{\left(\lambda \right)}_{2}D}{d})$ ....eq2
Both monochromatic lights have same fringe width area,
$60{\lambda }_1\frac{D}{d}=n{\lambda }_2\frac{D}{d}60\times 4000=n\times 6000n=40$