Question

A beam of light constituting of two wavelengths $6000\stackrel{\circ }{A}$ and $5200\stackrel{\circ }{A}$ is used to obtain interference fringes in a Young's double slit experiment. The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm. The least distance from the central maximum where the bright fringes due to both the wavelengths coincide is

• A. 3.056 cm
• B. 2.156 cm
• C. 0.156 cm
• D. 3.156 cm

Answer 1

The correct option is C 0.156 cm

${\beta }_1=\frac{{\lambda }_{1}D}{d};{\beta }_2=\frac{{\lambda }_{2}D}{d}$
$\frac{{\beta }_1}{{\beta }_2}=\frac{{\lambda }_1}{{\lambda }_2}=\frac{6500}{5200}=\frac{5}{4}$
or, $4{\beta }_1=5{\beta }_2$
$\Rightarrow$ Distance of 4th bright fringe due to ${\lambda }_1$
= Distance of 5th bright fringe due to ${\lambda }_2$
or, $x=4{\beta }_1=5{\beta }_2$
or, $x=\frac{4{\lambda }_{1}D}{d}=0.156cm$.