Conversation started Jul 1, 2020 at 6:12.
Guru Vishnu
Guru Vishnu
Jul 1, 2020 06:12
If a minima in YDSE occurs directly in front of one of the slits, then the wavelength of the radiation used is:
(a) 2 cm only
(b) 4 cm only
(c) 2 cm, 2/3 cm, 2/5 cm and so on
(d) 4 cm, 4/3 cm, 4/5 cm and so on
Given, D=12 cm and d=5 cm; where D and d are the distance between the slit plane and the screen and the distance between the two slits respectively.
I figured out the correct answer (c) by simply figuring out the path difference for a point opposite to a slit and applying the condition for minima. But I want to know why my alternate method as described below failed:
If a minima occurs opposite to one of the slits, the by symmetry it appears opposite to the other slit too. Now, points directly opposite to both the slits are minima. We know that the slit width is given by $w=D\lambda/d$. Using this and the fact that both the points directly opposite to the two slits are dark fringes, I arrived at the following:
$$\frac{nD\lambda}{d}=d$$
Where $n$ is an integer. However, on solving the above, I don't even get close to any of the options. So why does this method fail? I'm interested in knowing the reason this fails because this was the method which came to my mind first.
(End of message)
John Rennie
John Rennie
@GuruVishnu Where does the slit width come into this?
Guru Vishnu
Guru Vishnu
@JohnRennie Ah. Sorry sir, that was a typo. It's fringe width not slit width sir.
John Rennie
John Rennie
You mean the spacing between fringes?
Guru Vishnu
Guru Vishnu
Yes sir. It gives the distance between consecutive dark or bright fringes.
The equation I arrived at is simply based on fitting as many as bright fringes within the distance between the two slits on the screen, ensuring dark fringes always appear opposite to the slits.
John Rennie
John Rennie
The spacing between consecutive fringes is $D\lambda/d$, and as you argue, the spacing between fringes has to be equal to $d/n$ for some integer $n$ i.e. an integral number of fringe spacings equals $d$.
$$ \frac{D\lambda}{d} = \frac{d}{n} $$
Guru Vishnu
Guru Vishnu
Jul 1, 2020 06:23
Yes sir. Isn't that exactly the same equation I mentioned?
John Rennie
John Rennie
So I get the same result as you ...
Guru Vishnu
Guru Vishnu
Ok sir. So I guess there's no mistake in this argument.
Writing it for $\lambda$, we get:
$$\lambda=\frac{d^2}{Dn}$$
where we know the values of $d$ and $D$. But on substituting this and different values of $n$, we don't get any of the options, sir.
John Rennie
John Rennie
Ah wait, the equation $n\lambda = d\sin\theta$ only applies in the Fraunhofer limit.
Guru Vishnu
Guru Vishnu
Oh! Ok sir. So the equation for fringe width is invalid in this case as $d$ is not much smaller than $D$. Am I right sir?
D=12 and d=5; (both in centimetres)
John Rennie
John Rennie
In this case the slit spacing and screen distance are comparable so we'll be in the Fresnel region and the equation won't apply.
@GuruVishnu yes
Guru Vishnu
Guru Vishnu
Jul 1, 2020 06:27
Ok sir. Thank you for the clarification :-)
John Rennie
John Rennie
:-)
Guru Vishnu
Guru Vishnu
Otherwise, is this method logically right?
John Rennie
John Rennie
Yes, that's the way I immediately thought to do it.
Guru Vishnu
Guru Vishnu
Ok sir. Then that's a sign I've started to think like you :-)
John Rennie
John Rennie
@GuruVishnu whether that's a good thing is debatable :-)
 
Conversation ended Jul 1, 2020 at 6:32.