11:12 AM
If a minima in YDSE occurs directly in front of one of the slits, then the wavelength of the radiation used is:
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:
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.
If a minima in YDSE occurs directly in front of one of the slits, then the wavelength of the radiation used is:
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:
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1:18 PM
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