Question

A beam of light consisting of two wavelengths, 650 nm and 520 nm is used to obtain interference fringes in a Youngs 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)

1.56 mm

Let at liner distance 'y' from center of screen the bright fringes due to both wavelength coincides. Let the $n_1$ number of bright fringe with wevelength ${\lambda }_1$ coincides with $n_2$ number of bright fringe with wavelength ${\lambda }_2$.

We can write

$y=n_1{\beta }_1=n_2{\beta }_2$

$n_1\frac{{\lambda }_{1}D}{d}=n_2\frac{D{\lambda }_2}{d}or{n}_1{\lambda }_1=n_2{\lambda }_2$

Also at first position of coincide the ${}^{th}$ bright fringe of one will coincide with $(n+1{)}^{th}$ bright fringe of other.

If ${\lambda }_2<{\lambda }_1,$

So, then $n_2>n_1$ and $n_2=n_1+1$

Using equation (ii) in equation (i)

$n_1{\lambda }_1=(n_1+1){\lambda }_2$

$n_1\left(650\right)\times {10}^{-9}=(n_1+1)520\times {10}^{-9}$

$65n_1=52n_1+52or12n_1=52or{n}_1=4$

Thus, $y=n_1{\beta }_1=4\left[\frac{(6.5\times {10}^{-7})\left(1.2\right)}{2\times {10}^{-3}}\right]$

$=1.56\times {10}^{-.3}m=1.56mm$

So, the fourth bright fringe of wavelength 520 nm coincides with $5^{th}$ bright fringe of wavelength 650 nm.