Question

A beam of light consisting of two wavelengths, $650\,nm$ and $520 \,nm$ 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. $ 1.17\, mm$
• B. $ 2.52\, mm$
• C. $ 1.56\, mm$
• D. $ 3.14\, mm$

Answer 1

Answer: (C)

Let at linear distance $'y'$ 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
$n_1\lambda_1 = n_2\lambda_2\quad ...(i)$
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 $n_2 = n_1 + 1 \quad...(ii)$
Using equation $(ii)$ in equation $(i)$
$n_1\lambda_1 = (n_1 + 1) \lambda_2$
$n_1 (650) \times 10^{-9} = (n_1 + 1) 520 \times 10^{-9}$
$65 \,n_1 - 52 \,n_1 + 52$
or $13 \,n_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}}]$
$= 1.56 \times 10^{-3}\,m$
$ = 1.56\,mm$
So, the fourth bright fringe of wavelength $520\, nm$
coincides with $5^{th}$ bright fringe of wavelength $650 \,nm$