Question

In Young's double slit experiment, a minima is obtained when the phase difference of superimposing waves is:

• A. Zero
• B. $(2n-1)\pi$
• C. $n\pi$
• D. $(n+1)\pi$

Answer 1

The correct option is B $(2n-1)\pi$

For minima at a point the phase difference between the two waves reaching the point should be an odd integral multiple of $\pi$. In interference phenomenon, when both the waves are in opposite phase, we got destructive interference (or minima).
As resultant amplitude of two superimposed wave is
$R=\sqrt{a_1^2+a_2^2+2a_1{a}_{2}cos\varphi }$
and resultant intensity of two superimposed wave is
$I=I_1+I_2+2k{a}_1{a}_{2}cos\varphi$
For minima (destructive interference), phase difference should be odd multiple of $\pi$.
ie, $\varphi =(2n-1)\pi$
where, $n=1,2,3,...$
As $\varphi =\pi ,3\pi ,5\pi ,...$
Hence, $cos\varphi =cos(2n-1)\pi =-1$
So, $R_{min}=\sqrt{a_1^2+a_2^2-2a_1{a}_2}=\sqrt{(a_1-a_2{)}^2}$
or $R_{min}=a_1-a_2$
Also, $I=I_1+I_2-2k{a}_1{a}_2$
or $I=I_1+I_2-2a_1\sqrt{k}{a}_{\sqrt{k}}$
$=I_1+I_2-2\sqrt{I_1{I}_2}$
Hence, $I<I_1+I_2$
Hence, when the phase difference is odd multiple of $\pi$, then destructive interference is obtained. The resultant amplitude is equal to the difference of the individual amplitude and the resultant intensity is less than the sum of individual intensities.