Question

In a double slit experiment, the angular width of a fringe is found to be ${{0.2}^{0}}$ on a screen placed $1m$ away. The wavelength of light used is $600nm$. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take the refractive index of water to be $\dfrac{4}{3}$.

Answer 1

Hint: This problem can be solved by using the relation between the angular width, distance between the slits and the wavelength of the light. If the experimental apparatus is immersed in water, the wavelength of the light and hence, the angular width will change but the distance between the slits will remain constant as it is a physical characteristic of the setup.

Formula used:
The angular fringe width $\theta$ in a double slit experiment is given by

$\theta =\dfrac{\lambda }{d}$

where $\lambda$ is the wavelength of the light used and $d$ is the separation between the slits.
The wavelength of light $\lambda '$ in a medium is given by

$\lambda '=\dfrac{\lambda }{n}$

where $n$ is the refractive index of the medium and $\lambda$ is the wavelength of the light in vacuum.

Complete step by step answer:
In a double slit experiment, the angular width of a fringe depends upon the wavelength of the light used and the separation between the slits. When the apparatus is immersed in a different medium, the wavelength of the light will change but the separation between the slits will remain constant (since it is the physical characteristic of the setup and is independent of the medium in which it is placed). Therefore, the angular fringe width will also change accordingly.
The angular fringe width $\theta$ in a double slit experiment is given by

$\theta =\dfrac{\lambda }{d}$ --(1)

where $\lambda$ is the wavelength of the light used and $d$ is the separation between the slits.
The wavelength of light $\lambda '$ in a medium is given by

$\lambda '=\dfrac{\lambda }{n}$ --(2)
where $n$ is the refractive index of the medium and $\lambda$ is the wavelength of the light in vacuum.

Let us analyze the question.

Let the angular fringe width when the apparatus is in air be $\theta ={{0.2}^{0}}$.
The separation between the slits is $d$.
The wavelength of the light is $\lambda =600nm$.
Using (1), we get,

$d=\dfrac{\lambda }{\theta }$. --(3)
Now, the apparatus is immersed in water.

Given the refractive index of water is $n=\dfrac{4}{3}$.
Let the angular fringe width when the apparatus is in air be $\theta '$.
The separation between the slits is $d$.
The wavelength of the light is $\lambda '$.

Now using (2), we get,

$\lambda '=\dfrac{\lambda }{n}$ --(4)
Also, using (1), we get,
$d=\dfrac{\lambda '}{\theta '}$ --(5)
Therefore equating (3) and (5), we get,
$\dfrac{\lambda }{\theta }=\dfrac{\lambda '}{\theta '}$
$\therefore \dfrac{\lambda }{\theta }=\dfrac{\lambda }{n\theta '}$ [Using(4)]
$\therefore \theta '=\dfrac{\theta }{n}=\dfrac{{{0.2}^{0}}}{\dfrac{4}{3}}=\dfrac{3\times {{0.2}^{0}}}{4}={{0.15}^{0}}$

Hence, the angular fringe width in water will be ${{0.15}^{0}}$.

Note: Students should not get confused upon seeing that the distance of the screen is also given in the question and therefore has to be used somewhere. In fact, sometimes extra information is given in questions just to unnerve and confuse students. Upon solving the question students must have gotten a feel for the importance of the medium in the double slit experiment. This is mainly because the medium has an impact on the wavelength of light and the wavelength is one of the most important parameters in all wave properties. However, students must also remember that the medium does not have any effect on the frequency of the wave and it remains constant.