Question

In a Young’s double-slit experiment, the slits are separated by $0.28\text{mm}$ and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2\text{cm}$. Determine the wavelength of light used in the experiment.

Answer 1

Formula used: $x_n=\frac{n\lambda D}{d}$
Given,
Separation between the slits,
$d=0.28\text{mm}=0.28\times {10}^{-3}m$
Separation between the slits and the screen, $D=1.4m$
Distance between the central fringe and the fourth (n = 4) fringe,
$x_4=1.2\text{cm}=1.2\times {10}^{-2}m$

For constructive interference or bright fringes,
$x_n=\frac{n\lambda D}{d}[n=0,1,\mathrm{2....}]$
Where, $n$= order of fringes =4 and $\lambda =$
wavelength of light.

Putting the values we get,
$1.2\times {10}^{-2}=\frac{4\times \lambda \times 1.4}{0.28\times {10}^{-3}}$
$\lambda =\frac{1.2\times {10}^{-2}\times 0.28\times {10}^{-3}}{4\times 1.4}$
$=600\times {10}^{-9}m$
$=600\text{nm}$
Final Answer: $600\text{nm}$.