A question on Interference of light from two coherent sources?

Question

Two coherent sources A and B are d = 4 lemda meters apart, lemda being the wave-length of the light emitted by the sources.
An obsever goes arourd the two sources, along a circle of large radius D centered midway between A and B.
How many bright points would this observer come across in all as he completes the circular journey ?

Answer 1

Hello sir,

Let me first say that it's an honour to be answering your question. I would mention again that I fear I might be underqualified to be answering your question as you sir, hold a Ph.D. in physics. *Phew!*

Ok, here is my answer-

For bright fringes, we have the equation-

d*sinθ = mλ................................. (eq. 1)

Where,
d = distance between A and B
θ = Angle between the mid-point on AB and observer at distance D
λ = Wavelength of the light emitted by A and B.
m = an integer- 0, 1, 2, 3, 4.....

Now, as you have stated in the question-

d = 4λ ............................................ (eq. 2)

So, by eq. 1 and 2, we can say that-

4λ * sinθ = mλ

Or, sinθ = m/4................................ (eq. 3)

Now, for m= 0

sinθ = 0

i.e. θ = 0º (the central maximum)

And for m = 4

sinθ = 4/4 = 1

i.e. θ = 90º

Now, there are three integers in between 0 and 4 so, there would be three more maxima in addition to the two seen above at 0º and 90º as the observer moves to cover ¼th of the circular track.

Completing half of the circular track, all in all, the observer will encounter 9 bright fringes. (4 above and 4 below the central maximum).

Now, for the central maximum on the opposite side, we'll have a similar pattern of 4+1+4 = 9 bright fringes. Subtracting the 2 bright fringes we have already counted on the other side, we will get 7.

Now, adding the bright fringes obtained on both sides of the sources A and B-

(4+1+4) + (3+1+3) = 16.

So, the observer will come across total 16 bright fringes on his journey.

~IMPORTANT NOTE~

It'd be wrong to assume that the observer is identical to the Vertical Screen used in Young's double slit experiment as the path of the observer is clearly curved rather than vertical.

But, as the number of maxima and minima is not dependent on the distance of the screen from the source, we can safely assume the observer to be an "approaching vertical screen".

It'd have been a different matter altogether if we had to take the "position" of fringes in account too.

~ The End ~

I hope my answer was right.... but please be kind enough to point out any mistakes.

Thanks.

Answer 2

A cat's eyesight is 6 situations more effective perfect than a human in decrease lights contitions. even as taking off (or dilating), a cat's students can dilate a lot swifter than yours or mine. they could also dilate 3 situations more effective than yours or mine can. This shall we in diverse sunshine, and is between the justifications cats have their renowned capacity so see nicely in low mild. they don't see nicely in entire darkness, it really is a nicely-loved fantasy, in spite of the indisputable fact that it really is authentic that their eyes' capacity to dilate to this kind of wide degree is area of the reason they have an wonderful capacity to confirm in poorly lit situations. the only component everyone seems conscious about a cat's eyes are the students. A human student is continually round, yet a cat's student might want to be both round (even as that is dilated) or it may shrink in from the perimeters in stages till it really is elliptical. even as elliptical, a cat's eyes appear like a slit, stretching from good to bottom. This particular type facilitates a cat to squint its eyelids, masking in basic terms area of the student yet not all of it, subsequently giving it a minimum of a few guide administration over the quantity of sunshine enable in. If the student became in basic terms decreased to a small circle, last the eyelids would hide the student all at the same time.