Question

Two sources of light of wavelength $2500\stackrel{\circ }{A}$ and $3500\stackrel{\circ }{A}$ are used in Young’s double-slit experiment simultaneously. Which order of fringes of two-wavelength patterns coincide?
A. 3rd order of 1st source and 5th order of 2nd source.
B. 7th order of 1st source and 5th order of 2nd source.
C. 5th order of 1st source and 3rd order of 2nd source.
D. 5th order of 1st source and 7th order of 2nd source.

Answer 1

Hint: The difference between the paths in simple trigonometry can be written as $d\sin \theta$ , where d is the distance between the slits. The path length difference must be an integral multiple of the wavelength of the light source, or $d\sin \theta =n\lambda$ , for $n=0,1,-1,2,-2....$

Complete step by step answer:
Let us assume that $n_1$ fringe of source of wavelength ${\lambda }_1$ coincides with the $n_2$ fringe of source of wavelength ${\lambda }_2$ .
${\lambda }_1=2500\stackrel{\circ }{A}$
${\lambda }_2=3500\stackrel{\circ }{A}$
Using the formula , $d\sin \theta =n\lambda$
We get , $$n_1{\lambda }_1=n_2{\lambda }_2$$
$\Rightarrow {\displaystyle \frac{n_1}{n_2}}={\displaystyle \frac{{\lambda }_1}{{\lambda }_2}}$
After substituting the values of ${\lambda }_1$ and ${\lambda }_2$ in the above equation, we get
${\displaystyle \frac{n_1}{n_2}}={\displaystyle \frac{2500}{3500}}\Rightarrow {\displaystyle \frac{n_1}{n_2}}={\displaystyle \frac{5}{7}}$
Or ${\displaystyle \frac{n_1}{n_2}}={\displaystyle \frac{5k}{7k}}$ (where k is any integer)
By comparing the options we get $n_1$= 7 and $n_2$= 5
Hence, (b) option is the correct option.

Note: The equation $d\sin \theta =n\lambda$ (for $n=0,1,-1,2,-2....$) describes constructive interference. For fixed values of d and$\lambda$ , the larger n is, the larger $\sin \theta$ is. However, the maximum value that $\sin \theta$ can have is 1, for an angle of 90 degrees (Larger angle imply that light goes backward and does not reach the screen). Similarly for destructive interference for a double slit, the path length difference must be a half-integral multiple of the wavelength, or $d\sin \theta =\left(m+{\displaystyle \frac{1}{2}}\right)\lambda$ , for $m=0,1,-1,2,-2.....$