Animated Solution for Physics - Optics: In a double-slit experiment, green light ($5303\text{ A˚}$) falls on a double slit having a separation of $19.44\mu \text{-m}$ and a width of $4.05\mu \text{-m}$. The number of bright fringes between the first and the second diffraction minima is

Imagine a Young's double-slit experiment, but this time, let's strip away the idealizations. In textbook problems, we often treat slits as infinitely thin lines that produce perfectly uniform interference fringes. However, in reality, these slits have a finite width, denoted by $a$. Because of this real width, the beautiful interference fringes we usually see are actually trapped inside a larger, overarching envelope. This envelope is the diffraction pattern created by each individual slit.

To understand the final pattern on the screen, we must realize that the total intensity is the product of two phenomena: the broad diffraction envelope and the sharp interference fringes. The question asks us to look at a very specific region of this pattern—the space strictly between the first and second diffraction minima.

First, let's find where this diffraction envelope goes to zero. The condition for diffraction minima for a single slit of width $a$ is given by the equation:

Since the angles involved in these optical setups are extremely small, we can safely use the small-angle approximation, $\sin \theta \approx \theta$. This simplifies our condition to:

The question directs our attention to the region between the first and second diffraction minima. By plugging in $n=1$ and $n=2$, we can pinpoint the exact angular boundaries of this region. The first minimum occurs at ${\theta }_1=\frac{\lambda }{a}$, and the second minimum occurs at ${\theta }_2=\frac{2\lambda }{a}$.

Now, what about the bright fringes dancing inside this envelope? These are caused by the classic interference between the light waves emerging from the two separate slits, which are spaced a distance $d$ apart. The condition for an interference maximum (a bright fringe) is:

Again, applying the small-angle approximation, the angular position of the $m$-th bright fringe is:

We need to find exactly how many of these bright interference fringes fall strictly between our two diffraction boundaries. Mathematically, this is a beautiful constraint problem. We just set up a strict inequality:

Let's substitute the angular positions we just derived into this inequality:

Notice something elegant here? The wavelength $\lambda$ appears in every single term! This means the specific color of the light—whether it's the $5303\text{ A˚}$ green light given in the problem or any other wavelength—completely cancels out. The number of fringes is a purely geometric property of the slits. Multiplying the entire inequality by $d$, we get a remarkably simple relation:

Now, let's bring in our given physical dimensions. The slit separation $d$ is $19.44\mu \text{m}$, and the slit width $a$ is $4.05\mu \text{m}$. Let's calculate their crucial ratio:

Substituting this geometric ratio back into our master inequality, we get:

Since $m$ represents the order of the bright fringe (like the 1st, 2nd, or 3rd fringe), it must strictly be an integer. We need to find all the integers that lie strictly between $4.8$ and $9.6$. These integers are $5,6,7,8,$ and $9$.

Counting them up, we have exactly 5 bright fringes residing in this specific region.

As a fascinating side note, what if the ratio $d/a$ was exactly an integer, say $5$? Then the 5th interference maximum would fall exactly on the first diffraction minimum. The diffraction envelope would force the intensity to zero right where the interference peak is trying to form, causing the fringe to completely vanish! This phenomenon is known as a missing order, a critical concept to keep in your arsenal for advanced optics problems.