Question

In a Young's double slit experiment, slits are separated by $0.5mm$, and the screen is placed $150cm$ away. A beam of light consisting of two wavelengths, $650nm$ and $520nm$, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is

• A. $15.6mm$
• B. $1.56mm$
• C. $7.8mm$
• D. $9.75mm$

Answer 1

The correct option is C $7.8mm$

Fringe width of pattern due to ${\lambda }_1$ wavelength only $=\frac{{\lambda }_{1}D}{d}={\beta }_1$
Fringe width of pattern due to ${\lambda }_2$ only $=\frac{{\lambda }_{2}D}{d}={\beta }_2$

Hence, the distance from central maximum to point where the bright fringes first meet must be the least integral multiple common to both.
$n_1{\beta }_1=n_2{\beta }_2$
$\Rightarrow n_1\frac{{\lambda }_{1}D}{d}=n_2\frac{{\lambda }_{2}D}{d}$
$\Rightarrow n_1\left(650\right)=n_2\left(520\right)$
$\Rightarrow n_1=4,n_2=5$
Hence the required distance $=n_1\frac{{\lambda }_{1}D}{d}$
$=4\times \frac{(150\times {10}^{-2})\times (650\times {10}^{-9})}{(0.5\times {10}^{-3})}$
$=7.8mm$