Question

In Young’s double-slit experiment, the angular width of a fringe formed on a distant screen is ${{1}^{o}}$. The wavelength of light used is $6000{{A}^{o}}$. What is the spacing between the slits?
A.$344\,mm$
B.$0.1344\,mm$
C.$0.0344\,mm$
D.$0.034\,mm$

Answer 1

In this question, we have given the value of angular fringe width and wavelength. We are required to find the value of spacing between the slits and in order to do that we will apply the basic formula of an angular fringe width. In modern physics, 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.

Complete answer:
Given:
${{\beta }_{\theta }}={{1}^{o}}$
$\lambda =6000{{A}^{o}}$
$d=?$
Angular fringe width can be described as:
${{\beta }_{\theta }}=\dfrac{\lambda }{d}\,\,\,\,.....(1)$
Where,
${{\beta }_{\theta }}=$Angular fringe width
$\lambda =$ Wavelength
$d=$Spacing between the slits
The angular fringe width is the distance of the central fringe from the slit. It also indicates the angular separation between the fringes.
Putting the value of in equation (1), we get:
${{\beta }_{\theta }}=\dfrac{6000\times {{10}^{-10}}}{d}$
Therefore,
$d=\dfrac{6\times {{10}^{-7}}}{\dfrac{1\times \pi }{180}}=0.344\times {{10}^{-4}}m$
Or
$d=0.0344\,mm$
Therefore, the spacing between the slits is $0.0344\,mm$.

Hence option C. is correct.

Note:
Young’s double-slit experiment uses two coherent sources of light placed at a small distance apart, usually, only a few orders of magnitude greater than the wavelength of light is used. Young’s double-slit experiment helped in understanding the wave theory of light. The experiment belongs to a general class of "double path" experiments, in which a wave is split into two separate waves that later combine into a single wave. Changes in the path-lengths of both waves result in a phase shift, creating an interference pattern.