Question

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

Answer 1

Hint: Young’s double slit experiment was performed by Thomas Young in 1801 to understand the wave nature of light waves. It uses two light sources placed closer to each other at a small distance to show how interference of light takes place.

Formula Used:
Since bright fringes are being formed, constructive interference is taking place. Therefore, the formula used will be:
\[x = n\lambda \dfrac{D}{d}\]……(i)
Where D is the distance of the screen from the slits, ‘d’ is the distance between the slits, ‘n’ is the number of fringes and \[\lambda \] is the intensity.

Complete step by step solution:
It is given that the slits are separated by a distance , $d=0.28\,mm$.
Or it can be written that \[d = 0.28 \times {10^{ - 3}}\,m\].
Also, it is given that the slit and the screen is separated by a distance of \[D = 1.4\,m\].
The distance between the central and fourth fringe is given as $x=1.2\,cm$.
Or \[x = 1.2 \times {10^{ - 2}}\,m\]
Also, there are four fringes, therefore n=4.

Substituting the given values in equation (i) and solving we get
\[1.2 \times {10^{ - 2}} = 4\lambda \times \dfrac{{1.4}}{{0.28 \times {{10}^{ - 3}}}}\]
\[\Rightarrow \lambda = \dfrac{{1.2 \times {{10}^{ - 3}} \times 0.28 \times {{10}^{ - 3}}}}{{4 \times 1.4}}\]
\[\Rightarrow \lambda = 6 \times {10^{ - 7}}m\]
\[\therefore \lambda = 600\,nm\]

Therefore, the wavelength of the light that is used in the given experiment will be 600nm.

Note:In Young’s double slit experiment, there are two slits used. When the rays of light fall on these slits, they emit light waves forming crests and troughs to form a pattern. The patterns of these waves are formed such that they superimpose either constructively or destructively. If they superimpose constructively, a bright fringe or band of high intensity is formed. If they superimpose destructively, then a dark fringe or a dark band of low intensity is formed. These patterns that are formed due to bright or dark fringes are known as interference patterns.