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.(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength $650nm$.(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the two slits is $0.28mm$ and the screen is at a distance of $1.4m$ from the slits.

Answer 1

Given

The wavelength of the light is ${\lambda }_1=650nm$

The wavelength of seond light,${\lambda }_2=520nm$

Distance between the slit and the screen is $1.4m$.

Distance between the slits is $0.28mm$.

$\left(a\right)$

The relation between the $n^{th}$ bright fringe and the width of fringe is:

$x=n{\lambda }_1\frac{D}{d}$

For third bright fringe, $n=3$

$x=3\times 650\frac{1.4}{0.28\times {10}^{-3}}=1950\times 5\times {10}^{3}nm$

$x=9.75\times {10}^{-3}m$

$=9.75mm$

$\left(b\right)$

We can consider that $n^{th}$ bright frind of ${\lambda }_2$ and the $(n-1{)}^{th}$ bright fringe of wavelength ${\lambda }_1$ coincide with each other.

$n{\lambda }_2=(n-1){\lambda }_1$

$520n=650n-650$

$650=130n$

$n=5$

therefore, the least distance from the central maximum can be obtained as:

$x^{\prime }=n{\lambda }_2\frac{D}{d}$

$x^{\prime }=5\times 520\frac{D}{d}=2600\frac{1.4}{0.28\times {10}^{-3}}nm$

$x^{\prime }=1.30\times {10}^{-2}m=1.3cm$