Question

In a Young's double slit experiment, slits are separated by $0.5 \, mm$ , and the screen is placed $150 \, cm$ away. A beam of light consisting of two wavelengths, $650 \, nm$ and $520 \, nm$ , 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 $n_{2}$
$n_{1}=4$ $n_{2}=5$
$\left(y_{1}\right)_{m i n}=4\times 650\times nm\times \frac{150 m}{0.5 \times \left(10\right)^{- 3} m}$
$=7800\times 10^{- 6}m$
$=7.8 \, mm$