Question

A beam of light consisting of two wavelengths, $650nm$ and $520nm$ is used to obtain interference fringes in a Young’s double-slit experiment. What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

• A. $1.17mm$
• B. $2.52mm$
• C. $1.56mm$
• D. $3.14mm$

Answer 1

Answer: (C)

Let at linear distance ${}^{\prime }{y}^{\prime }$ from center of screen the bright fringes due to both wavelength coincides. Let $n_1$ number of bright fringe with wavelength ${\lambda }_1$ coincides with $n_2$ number of bright fringe with wavelength ${\lambda }_2$.
We can write
$y=n_1{\beta }_1=n_2{\beta }_2$
$n_1\frac{{\lambda }_{1}D}{d}=n_2\frac{D{\lambda }_2}{d}$ or
$n_1{\lambda }_1=n_2{\lambda }_2...(i)$
Also at first position of coincide, the $n^{th}$ bright fringe of one will coincide with $(n+1{)}^{th}$ bright fringe of other.
lf ${\lambda }_2<{\lambda }_1$,
So, then $n_2>n_1$ and $n_2=n_1+1...(ii)$
Using equation $(ii)$ in equation $(i)$
$n_1{\lambda }_1=(n_1+1){\lambda }_2$
$n_1(650)\times 10^{-9}=(n_1+1)520\times 10^{-9}$
$65n_1-52n_1+52$
or $13n_1=52$ or $n_1=4$
Thus, $y=n_1{\beta }_1$
$=4[\frac{(6.5\times 10^{-7})(1.2)}{2\times 10^{-3}}]$
$=1.56\times 10^{-3}m$
$=1.56mm$
So, the fourth bright fringe of wavelength $520nm$
coincides with $5^{th}$ bright fringe of wavelength $650nm$