Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

Minimum value of d/$\lambda$ for destructive interference

1 vote
172 views

Three identical monochromatic points sources of light emit light of wavelength $\lambda$ coherently and in phase with each other. They are placed on the x-axis at the points x = – d, 0 and d. find the minimum value of d/$\lambda$ for which there is destructive interference with almost zero resultant intensity at points on the x-axis having x > > d.

I am little weak in this topic
I tried as
For destructive interfernce path difference should be $\lambda /2$
Let the destructive point would be at a distance x from origin , then
The path covered by each source would be x+2d , x+d and x.

asked Feb 10, 2017 in Physics Problems by koolman (4,286 points)
Answer is given as 1/3

1 Answer

1 vote
 
Best answer

For destructrive interference of TWO waves the path difference should be $\lambda/2$. But this doesn't apply for THREE waves.

The amplitude of the waves decreases as $1/x^2$. When $x \gg d$ the amplitudes will be approximately equal.

If the phase difference is $\phi=2\pi \frac{d}{\lambda}$ then the sum of the waves is zero when :
$\cos(-1\phi)+\cos(0\phi)+\cos(+1\phi) = 1 + 2\cos\phi=0$
because $\cos(-\phi)=\cos(+\phi)$.

Solve to find $\phi$.

answered Feb 10, 2017 by sammy gerbil (28,876 points)
edited Feb 12, 2017 by sammy gerbil
How you got the angles as -1$\phi$,0$\phi$,1$\phi$
*The difference in phase at point $x \gg d$ between adjacent sources is $\phi$. So when the phase of the middle source is $0$ the phase of the source at $-d$ is $-\phi$ and the phase of the source at $+d$ is $+\phi$.
Is it necessary when we are given x>>d
The amplitude of the waves is proportional to $1/x^2$. The condition $x \gg d$ ensures that the amplitudes are approximately the same, because then $1/(x-d)^2 \approx 1/x^2 \approx 1/(x+d)^2$. Then only the phase difference between the waves decides whether their interference results in complete cancellation.

However, even with $x \approx d$ it might be possible - but more difficult - to find values of $x$ and $\phi$ for which there is complete destructive interference - ie  
$\frac{A}{(x+d)^2}\cos(-\phi)+\frac{A}{x^2}\cos(0\phi)+\frac{A}{(x+d)^2}\cos(+\phi)=0$.

It is also possible to have different phase differences between sources 1-2 and 2-3.
...