Question

The ratio of intensity at the center of bright fringe to the intensity at a point distant one-fourth of the distance between two successive bright fringes will be:
A. $4$
B. $3$
C. $2$
D. $1$

Answer 1

Hint:To answer this problem, Young’s double-slit experiment procedure must be recollected and the mathematical calculations of ratio for bright and dark fringes are utilized to find their separate intensities. Hence from this concept, we can find the required answer in the given options.

Complete step by step solution:
Young’s double-slit experiment utilizes Monochromatic or Coherent sources. Here Monochromatic sources mean that they possess frequencies, but those frequencies are constant with time and coherent sources mean both the frequencies are equal, and also the phase difference is not changing concerning the time when there is any phase difference. The total intensity for maximum and minimum intensities should be done sensibly. Also, take the value of $\cos \phi $ properly.
 To solve this problem, we are going to use the theory of intensity distribution in Young's double-slit experiment, theory of path difference, and phase difference. When the intensity of the central bright fringe is ${I_o}$ , the intensity at a point is obtained from the formula, $I = {I_o}{\cos ^2}\left( {\dfrac{{\Delta \phi }}{2}} \right)$, where $\Delta \phi $ is the phase difference between the two waves.
So $I = {I_o}{\cos ^2}\left( {\dfrac{{\Delta \phi }}{2}} \right)$
$\Delta \phi $ at one-fourth of the distance between two successive bright fringes is given by
$\dfrac{{2\pi }}{4} = \dfrac{\pi }{2}$
So, $I = {I_o}{\cos ^2}\left( {\dfrac{{\pi /2}}{2}} \right)$
$ = {I_o}{\cos ^2}\left( {\dfrac{\pi }{4}} \right)$
\[ = {I_o}\dfrac{1}{{{{\left( {\sqrt 2 } \right)}^2}}}\] as $\cos (\pi /4) = \dfrac{1}{{\sqrt 2 }}$
$ = {I_o} \dfrac{1}{{2}} $
So the ratio of their intensities will be
$\dfrac{I_o}{{I}} = \dfrac{I_o}{{\dfrac{I_o}{{2}}}} = \dfrac{2}{1} = 2$
Hence Option C is correct.

Note:
The path difference between the two waves is given to be $\dfrac{1}{8}th$ of the wavelength of the interfering waves. It can be taken as, $\Delta x = \dfrac{\lambda }{8}$ where $\lambda $ is given as the wavelength of the interfering waves. To find the intensity at the given point with respect to the intensity of the central bright fringe, we need to estimate the phase difference. It can be given by applying the expression between path difference and phase difference, which is $\Delta \phi = \dfrac{{2\pi \Delta x}}{\lambda }$.