Question

In Young's double slit experiment the separation d between the slits is 2mm, the wavelength λ of the light used is 5896Å and distance D between the screen and slits is 100cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same λ and D) the separation between the slits needs to be changed to: -
(1) 1.8mm
(2) 1.9mm
(3) 2.1mm
(4) 1.7mm

Answer 1

Hint: In Young’s double-slit experiment interference of light takes place and due to constructive and destructive interference of light. The frequency of light and the phase difference between the two interfering waves determine the intensity and the type of pattern obtained on the screen.

Complete step by step answer:
Given d= 2 mm = \[2\times {{10}^{-3}}m\], D= 100 cm= 1 m, \[\lambda =5896{{A}^{0}}=5896\times {{10}^{-10}}m\], \[{{\theta }_{1}}\]=\[{{0.20}^{0}}\] and \[{{\theta }_{2}}={{0.21}^{0}}\]
We know angular width is given by the relation, \[\theta =\dfrac{\lambda }{d}\]
here \[\lambda \]is the wavelength of the light and d, is the separation between the two slits.
So, \[\dfrac{{{\theta }_{1}}}{{{\theta }_{2}}}=\dfrac{{{d}_{2}}}{{{d}_{1}}}\], since wavelength remains same. Putting the values, we get,
\[\dfrac{0.20}{0.21}=\dfrac{{{d}_{2}}}{2}\]
\[{{d}_{2}}=1.9mm\].

So, the correct answer is “Option 2”.

Note:
Interference of light is a phenomenon in which two or more waves interfere with each other and there is redistribution of energy. We can say two waves superpose to form a resultant wave of greater, lower, or the same amplitude. when an interference pattern is observed on the screen one distinct feature is all the bright bands are equally bright and all the dark bands are equally dark. While in case of difdfraction when the pattern is observed on the screen the bright bands are of decreasing intensity as we move away from the central maxima. This is a very clear distinction between the two patterns.