Young's Double Slit Experiment

[Priyadarshini]

When the slits are made narrower (but with same separation) why are more fringes produced? If the slits are narrower, less light enters, so less light interferes with each other, so lesser number of fringes should be produced, isn't it?

 

[blue_leaf77]

If the slits are narrower, less light enters, so less light interferes with each other, so lesser number of fringes should be produced, isn't it?

That's not how the logic goes for this problem. It's true that as the slit width decreases, the diffraction pattern becomes dimmer, but that's not the reason why there are more fringes observed within the central lobe of the diffraction pattern. In order to understand the reason, let's take a look at the equation of the diffraction pattern
$$
I(x) \propto \cos^2\left(\frac{\pi d x}{\lambda L}\right) \textrm{sinc}^2 \left(\frac{\pi b x}{\lambda L}\right)
$$
where $d$ the slit separation, $b$ slit width, and $L$ the distance between slit plane and the screen. In that equation, the fringes are described by the cosine term enveloped by a wider $\textrm{sinc}$ function. The width of the central lobe is proportional to $\frac{\lambda L}{\pi b}$. Therefore if the slit width $b$ decreases, the central lobe in the diffraction pattern becomes wider and more fringes are covered .

 

[Priyadarshini]

That's not how the logic goes for this problem. It's true that as the slit width decreases, the diffraction pattern becomes dimmer, but that's not the reason why there are more fringes observed within the central lobe of the diffraction pattern. In order to understand the reason, let's take a look at the equation of the diffraction pattern
$$
I(x) \propto \cos^2\left(\frac{\pi d x}{\lambda L}\right) \textrm{sinc}^2 \left(\frac{\pi b x}{\lambda L}\right)
$$
where $d$ the slit separation, $b$ slit width, and $L$ the distance between slit plane and the screen. In that equation, the fringes are described by the cosine term enveloped by a wider $\textrm{sinc}$ function. The width of the central lobe is proportional to $\frac{\lambda L}{\pi b}$. Therefore if the slit width $b$ decreases, the central lobe in the diffraction pattern becomes wider and more fringes are covered .

If the central lobe becomes wider, how are more fringes formed?
Is there a simpler logic? We haven't studied that formula yet.

 

[Charles Link]

If the central lobe becomes wider, how are more fringes formed?
Is there a simpler logic? We haven't studied that formula yet.

The first part of the intensity formula is the interference of two slits. The second part of the formula, which determines how wide the observed interference pattern is, is the diffraction pattern of a single slit. Ideally the slits are very narrow, making the diffraction pattern of a single slit, and thereby the entire pattern, very wide, but the slits need to be wide enough to let enough light through to see the pattern. As the slits are widened, the observed interference pattern narrows. (The single slit diffraction pattern has zero intensity at m*lambda=b*sin(theta) where m=non-zero integer. The central lobe (brightest area) is the region around m=0. Notice as the slit width b is made larger, the spread of angle theta between the m=1 and m=-1 zero's of the single slit diffraction pattern decreases. )

 

[Khashishi]

The diffraction pattern is the product of two effects: the single slit diffraction pattern and the idealized (zero width) double slit pattern. With an idealized double slit, all your fringes are the same brightness and you have unlimited number of them. If you've studied the single slit pattern, you should know that a smaller slit gives you a wider pattern.

 

[blue_leaf77]

Is there a simpler logic? We haven't studied that formula yet.

I believe the previous two posts have elaborated the answer in a simpler way. The quickest way, however, to get a grasp of how that equation behaves is to use a function plotter program to plot it and play around with the parameter $b$ (slit width).