Question

In YSDE, an angular position of the point on the central maximum whose intensity is $\dfrac{{\text{1}}}{{\text{4}}}{\text{th}}$of the maximum intensity.
A. ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{\lambda }{d}} \right)$
B. ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{\lambda }{{2d}}} \right)$
C. ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{\lambda }{{3d}}} \right)$
D. ${\text{si}}{{\text{n}}^{{\text{ - 1}}}}\left( {\dfrac{\lambda }{{4d}}} \right)$

Answer 1

From definition we know that at a phase difference of $\phi$ , resultant intensity is $I = 4{I_0}\,{\cos ^2}\left( {\dfrac{\phi }{2}} \right)$ then after calculation we find the Phase difference in YDSE.

Complete step by step solution:
Here,
From the definition we know that,
Phase Difference is used when two or more opposite quantities approach their limit or zero values to define the difference in degrees or radians.
At a phase difference of$\phi$, resultant intensity is:
$I = 4{I_0}\,{\cos ^2}\left( {\dfrac{\phi }{2}} \right)$
Given,
$\dfrac{I}{{{I_{\max }}}} = \dfrac{1}{4}$
$\Rightarrow \dfrac{{4{I_0}\,{{\cos }^2}\left( {\dfrac{\phi }{2}} \right)}}{{4{I_0}}} = \dfrac{1}{4}$
Therefore,
$\dfrac{\phi }{2} = \pm \dfrac{\pi }{3} + 2n\pi ,\, \pm \dfrac{{2\pi }}{3} + 2n\pi$
$\phi = \pm \dfrac{{2\pi }}{3} + 4n\pi ,\, \pm \dfrac{{4\pi }}{3} + 4n\pi$
Now,
Phase difference in YDSE is given by:
$\phi = \dfrac{{2\pi d\,\sin \theta }}{\lambda }$
From above equation, considering only central fringe, we get

$$\dfrac{{2\pi d\,\sin \theta }}{\lambda } = \pm \dfrac{{2\pi }}{3} \\\ \theta = \pm {\sin ^{ - 1}}\left( {\dfrac{\lambda }{{3d}}} \right) \\\$$

**Hence,
In YSDE, an angular position of the point on the central maximum whose intensity is $\dfrac{{\text{1}}}{{\text{4}}}{\text{th}}$of the maximum intensity is ${\sin ^{ - 1}}\left( {\dfrac{\lambda }{{3d}}} \right)$ **

Note: From definition we know that Phase Difference is used when two or more opposite quantities approach their limit or zero values to define the difference in degrees or radians.
Intensity of radiant energy is the power transferred per unit area, where the area on the plane is measured perpendicular to the direction of the energy propagation.
Young's double-slit experiment: When a distant screen illuminates monochromatic light which passes through two narrow slits, a characteristic pattern of bright and dark fringes is observed. This pattern of interference is caused by overlapping light waves which originate from the two slits. Constructive interference regions, corresponding to bright fringes, are created when the difference in path from the two slits to the fringe is an integral number of light's wavelength. Destructive interference and dark fringes arise when the difference in direction is a half-integrated number of wavelengths.