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)

Given,
Wavelength of first light is, ${\lambda }_1=650nm$
Wavelength of second light is, ${\lambda }_2=520nm$
Screen distance is, $D=150cm$
Slit width is, $d=0.5mm$
For first wavelength, location of fringe from central line,
$y_1=\frac{n_1{\lambda }_{1}D}{d}$ for bright fringes
For second wavelength, location of fringe from central line,
$y_2=\frac{n_2{\lambda }_{2}D}{d}$ for bright fringes
Condition of coincide of fringe is,
$y_1=y_2$
$\Rightarrow n_1{\lambda }_1=n_2{\lambda }_2$
$\Rightarrow n_1\times 650=n_2\times 520$
$\Rightarrow \frac{n_1}{n_2}=\frac{4}{5}$
For the minimum value of $n_1$ and $n_2$
$n_1=4$ and $n_2=5$
${\left(y_1\right)}_{min}=4\times 650nm\times \frac{1.50m}{0.5\times {\left(10\right)}^{-3}m}$
$=7800\times 10^{-6}m$
$=7.8mm$