Question

In a double slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. The angular width of the fringe if entire experimental apparatus is immersed in water is (Take ${\mu _{{\text{water }}}} = {\text{ }}\dfrac{4}{3}$)
$
  {\text{A}}{\text{. 0}}{\text{.15}}^\circ \\
  {\text{B}}{\text{. 1}}^\circ \\
  {\text{C}}{\text{. 2}}^\circ \\
  {\text{D}}{\text{. 0}}{\text{.3}}^\circ \\
$

Answer 1

Hint – To find the angular width of the fringe in water, we apply the formula of double slit experiment in both water and air medium, establish a relation between them using the given data and solve. The angular width when immersed in water is definitely different because the refractive index of air and water are different.
Formula Used: $\theta {\text{ = }}\dfrac{\lambda }{{\text{d}}}$
Where θ is the angular fringe separation, λ is the wavelength of the incident light and d is the width of the slit.

Complete step-by-step solution -
Given Data,
Angular width of the fringe = 0.2°
Wavelength of light used = 600 nm
Refractive index of water, ${\mu _{{\text{water }}}} = {\text{ }}\dfrac{4}{3}$
We know Angular Fringe separation in air medium is given by
$ \theta {\text{ = }}\dfrac{\lambda }{{\text{d}}} \\
   \Rightarrow {\text{d = }}\dfrac{{{\lambda _{\text{a}}}}}{{{\theta _{\text{a}}}}} \\ $
Let’s calculate when the apparatus is immersed in water,
The width of the slit ‘d’ is the same in air and in water. Let us consider the wavelength and the angular width of the fringe in water are ${\lambda _{\text{w}}}{\text{ and }}{\theta _{\text{w}}}$ .
Hence here ${\text{d = }}\dfrac{{{\lambda _{\text{w}}}}}{{{\theta _{\text{w}}}}}$.
As the distance d is the same, we equate them
$ \Rightarrow \dfrac{{{\lambda _{\text{a}}}}}{{{\theta _{\text{a}}}}} = \dfrac{{{\lambda _{\text{w}}}}}{{{\theta _{\text{w}}}}}$
$ \Rightarrow \dfrac{{{\lambda _{\text{a}}}}}{{{\lambda _{\text{w}}}}} = \dfrac{{{\theta _{\text{a}}}}}{{{\theta _{\text{w}}}}}$ --- (1)
Now we know refractive index of a medium is defined w.r.t the value of it in air medium, its formula is
$\left[ {{\mu _{\text{w}}} = {}_{\text{a}}{\mu _{\text{w}}} = \dfrac{{{\lambda _{\text{a}}}}}{{{\lambda _{\text{w}}}}}} \right]$.
Given ${\mu _{{\text{water }}}} = {\text{ }}\dfrac{4}{3}$. Substituting it in Equation (1) we get
$ \Rightarrow \dfrac{{{\theta _{\text{a}}}}}{{{\theta _{\text{w}}}}} = \dfrac{4}{3}$
\[ \Rightarrow {\theta _{\text{w}}} = \dfrac{3}{4}{\theta _{\text{a}}} \\
   \Rightarrow {\theta _{\text{w}}} = \dfrac{3}{4}\left( {0.2} \right) = 0.15^\circ \\
 \] -------- --- Given ${\theta _{\text{a}}} = 0.2^\circ $
Hence Option A is the correct answer.

Note – In order to answer this type of question the key is to know the meaning and the relevant formulae of a double slit experiment. It was performed by Thomas Young in the 1800’s. Refractive index of a material is a dimensionless number that describes how fast light travels through the material, it does not have any units.

The double-slit experiment is a demonstration that light and matter can display characteristics of both classically defined waves and particles; moreover, it displays the fundamentally probabilistic nature of quantum mechanical phenomena.