Question

The angular width of Central maxima of a diffraction pattern of a single slit does not depend upon
(A) Distance between slit and source
(B) Wavelength of light used
(C) Width of slit
(D) Frequency of light used

Answer 1

Hint: The diffraction pattern which is observed when light passes through a single slit and diffraction occurs. We know the formula of angular width in case of single slit diffraction and from that we can neglect the quantity on which angular width does not depend.

Complete Step-By-Step Solution:
We know that diffraction of light is defined as the spreading or scattering of light waves when it encounters a slit or an opening.
When this light wave passes through a slit such that the width of the slit is in the order of the wavelength of the light, we say single slit displacement has occurred.
We know that in case of single slit diffraction, the point at which the reinforcement of secondary waves occurs and the intensity of light is maximum is known as the Central maxima of the diffraction pattern.
We know that the angular width of central maxima is always in the order of the wavelength.
Now, the expression of angular width is given by:
$e\sin \theta = \lambda $
Or we can write:
\[\sin \theta = \dfrac{\lambda }{e}\]
\[ \Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{\lambda }{e}} \right)\]
Where,
\[\theta = \]Angular width
\[\lambda = \] Wavelength of light
\[e = \] Slit width
Thus, from the expression it is clear that the angular width depends upon the wavelength of the light and width of the slit.

Therefore, we can say among the options given, the angular width does not depend upon the distance between sources and slit, hence option (A) is correct.

Note:
We know that the wavelength of the light depends upon its frequency. Thus we can say that angular width also depends upon the frequency of the light.