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

Answer: (C)

$y_1=\frac{n_1{\lambda }_{1}D}{d}$ for bright fringes
$y_2=\frac{n_2{\lambda }_{2}D}{d}$ for bright fringes
To coincide
$n_1{\lambda }_1=n_2{\lambda }_2$
$n_1\times 650=n_2\times 520$
$\therefore \frac{n_1}{n_2}=\frac{4}{5}$
For minimum value of $n_1$ and $n_2$
$n_1=4$ $n_2=5$
${\left(y_1\right)}_{min}=4\times 650\times nm\times \frac{150m}{0.5\times {\left(10\right)}^{-3}m}$
$=7800\times 10^{-6}m$
$=7.8mm$