Question

In a Youngs$^,$s double slit experiment , the intensity at the central maximum is ${I_0}$ The intensity at a distance $\dfrac{\beta }{2}$ above the central maximum is (where $\beta $ = fringe width).

Answer 1

Hint: According to given question we have to find out the path difference and phase at a distance of $\dfrac{\beta }{2}$ unit from central maxima .After finding these values we have to apply maximum intensity equation to find two solvable equation. After solving we get the value and finally we get the answer of the question.

Complete step by step solution:
Here,
Central maxima intensity is ${I_0}$
As we already knows that $\beta = \dfrac{{D\lambda }}{d}$ where $\beta = $fringe width
                                                                           $D = $ separation between slits and screen
                                                                            $\lambda = $wavelength of light used
                                                                            $d = $separation between two slits
And also $y = \dfrac{{\lambda d}}{{2D}}$ where y is difference from center to maxima exceeds by $\dfrac{\beta }{2}$ unit
                                        And other conventions are same as above
Now, we know that path difference is equal to:
$\Delta x = \dfrac{{yd}}{D}$ where $\Delta x = $path difference
Now putting value of y in path difference equation, and after solving
$
  \Delta x = \dfrac{{\lambda d}}{{2D}} \times \dfrac{D}{d} \\
  \Delta x = \dfrac{y}{2} \\
 $
We have $\Phi = \dfrac{{2\pi }}{\lambda } \times \Delta x$ where $\Phi = $phase difference
After putting value of $\Delta x$ in phase difference equation:
$\Phi = \dfrac{{2\pi }}{\lambda } \times \dfrac{\lambda }{2}$
$\Phi = \pi $
Now apply maximum intensity equation:
${I_{\max }} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \Phi $
Where ${I_1}$ and ${I_2}$ are intensity from two slits
Now for central bright fringe
Let ${I_1} = {I_2} = {I_s}$ and
We also know that for central bright fringe
Hence ,
On further solving,
${I_o} = 4{I_s}$
Now apply maximum intensity equation for fringe above to central bright fringe
So, $I = {I_s} + {I_s} + 2\sqrt {{I_s}{I_s}} cos\pi $
On further solving $I = 0$

Hence, we can say that intensity of light at a distance $\dfrac{\beta }{2}$ unit above central maxima is 0.

Note:In young's double slit experiment we use two parallel slits and screen to get the fringes pattern.In these types of questions we talked about the phase difference of light waves and the path difference of light waves. And a maximum intensity equation to get the intensity at a particular distance from the central maxima.