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intensity of wave and wavelegth of light relation

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If $\lambda $ is the wavelength of ligth used and the distanse between the slits in Youngs double slit experiment is d then comment on number of fringes and fringe pattern if ,
1. d is 0.99 $\lambda $
2. $\lambda < d<2 \lambda $
Also the intensity of light falling on 1 slit is 4 times that of other

what I thought :

In a Young double-slit experiment why is there only one maxima formed if distance between the slits approaches wavelength of light.
what I understand so far that
$$I= I_1 +I_2 +2\sqrt {I _1I_2}cos\phi ,$$
where $I $ is resultant intensity
$I_1$ and $I_2$ are intensity of both waves $\phi $ is the phase difference
but then if so then,
how would intensity and the number of fringe pattern vary (even mathematically) because $\phi $=$2\pi \delta x/\lambda$ where $ \delta x$ is the path difference.

So how would that affect intensity and number of fringes?

What I know so far is that if the intensities of both the lights used are same then the contrast would be best and we would be able to differentiate fringes properly.

But I am unable to think about the wavelength and intensity part.

asked Feb 26, 2017 in Physics Problems by physicsapproval (2,320 points)
edited Feb 27, 2017 by physicsapproval

1 Answer

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Best answer

The question does not ask about intensity. You are looking at the wrong equation. You need an equation which includes the order $m$ of the fringes. The highest order $m$ with an angle of less than 90 degrees will give you the answer. The equation to use is

$m\lambda=d\sin\theta$

The question asks you to comment on the number of fringes and fringe pattern. The latter is rather too vague, I don't know what it means. Brightness perhaps? That is where your eqn might get used. Sharpness? Sharpness depends on the width $w$ of the slits compared with their separation $d$ - but we are not given $w$.

answered Feb 26, 2017 by sammy gerbil (27,948 points)
selected Mar 24, 2017 by physicsapproval
answer says if d=0.99 $\lambda $ the screen will contain only one maximum.
And for second case : atleast one more maximum (besides the central maximum ) will be observed on the screen
Also the intensity of light falling on 1 slit is 4 times that of other.. I forgot to mention that in the question , sorry for that , but will that make make any diffrence?
I didn't understand the ise of the equation mλ=2dsinθ given by you as isn't this the Bragg's law for X-rays while determining the interatomic distances? Can we use it here too ?
I wrote the equation wrongly. I have corrected it now.  

Intensity does not affect the number of fringes, only their visibility - ie how easy it is to see them.
Still I don't understand why only one  maximum  at 0.99$\lambda $
Insert $d=0.99\lambda$ and $d=\lambda, 2\lambda$ into the equation. How many values of $m$ allow you to find a solution for $\theta$? Recall that $\sin\theta$ has a maximum value of 1. There will always be an $m=0$ fringe; after that one each side for each $m\ge 1$.
yes , got that , thanks
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