A beam of light contains two wavelengths 6500 $\stackrel{0}{\mathrm{A}}$ and 5200 $\stackrel{0}{\mathrm{A}}$. They form interference fringes in Young's double slit experiment. What is the least distance (approx) from central maximum, where the bright fringes due to both wavelength coincide? The distance between the slits is 2 mm and distance between the screen and slit is 120 cm. [CG PMT 2015]

To find the point of coincidence of bright fringes, we can equate the distance of bright fringes from the central maximum, made by both the wavelengths of light. 

Given, distance between the screen and slit,
D = 120 cm = 120 x 10^-2 m
Slit width, d = 2 mm = 2 x 10^-3 m
$\lambda$₁ = 6500 $\stackrel{0}{\mathrm{A}}$ and $\lambda$₂ = 5200 $\stackrel{0}{\mathrm{A}}$
Let nth fringe due to  $\lambda$₂ = 5200 $\stackrel{0}{\mathrm{A}}$ coincide with (n $-$1) th
bright fringe due to  $\lambda$₁ = 6500 $\stackrel{0}{\mathrm{A}}$.

$\therefore \frac{Dn{\lambda }_2}{d}=\frac{D(n-1){\lambda }_1}{d}$

or    n  x  5200 = (n $-$ 1) 6500
or   4n = 5n $-$ 5

$\Rightarrow$  n = 5
So,  the least distance required,  $y_n=\frac{n{\lambda }_{2}D}{d}$

$=\frac{5\times 5200\times {10}^{-10}\times 120\times {10}^{-2}}{2\times {10}^{-3}}$

$=0.16\times {10}^{-2} \mathrm{m}=0.16 \mathrm{cm}$