Animated Solution for Physics - Optics: In a Young's double slit experiment with slit separation $0.1\text{ mm}$, one observes a bright fringe at angle $\frac{1}{40}\text{ rad}$ by using light of wavelength ${\lambda }_1$. When the light of the wavelength ${\lambda }_2$ is used a bright fringe is seen at the same angle in the same set up. Given that ${\lambda }_1$ and ${\lambda }_2$ are in visible range ($380\text{ n-m}$ to $740\text{ n-m}$), their values are

Have you ever wondered what happens when you shine two different colors of light through a double-slit apparatus? The screen doesn't just show a simple pattern; it becomes a complex canvas of overlapping waves. In this problem, we are tasked with playing detective. We know exactly where a bright fringe appears, and we need to figure out which colors of light could have created it.

Let's embark on this journey through the physics of interference.

Imagine the classic Young's Double Slit Experiment. We have two tiny slits, $S_1$ and $S_2$, separated by a microscopic distance $d=0.1\text{ mm}$. Light passes through these slits and travels to a distant screen. We are focusing our attention on a very specific point $P$ on that screen, located at an angular position of $\theta =\frac{1}{40}\text{ rad}$.

Why does a bright fringe form at $P$? It all comes down to the path difference ($\Delta x$). The light ray traveling from the bottom slit $S_2$ has to cover a slightly longer distance than the ray from the top slit $S_1$. Geometrically, if we drop a perpendicular from $S_1$ to the second ray, this extra distance is given by:

Here is where we use a powerful physicist's tool: the small-angle approximation. Because $\theta =\frac{1}{40}\text{ rad}$ is extremely small (roughly $1.4^{\circ }$), the sine of the angle is practically identical to the angle itself in radians. Therefore, we can safely write:

Let's plug in our numbers. We must be careful with units, converting our slit separation $d$ into meters ($10^{-4}\text{ m}$):

To make this number easier to work with, let's convert it to nanometers ($\text{nm}$), the standard unit for visible light wavelengths. Multiplying by $10^9$, we get a path difference of exactly $2500\text{ nm}$.

Now, we bring in the core principle of wave optics. For a bright fringe to appear at point $P$, the two light waves must arrive perfectly in phase. This constructive interference only happens when the path difference is an exact integer multiple of the light's wavelength $\lambda$.

This gives us our master equation:

Where $n$ is an integer ($1,2,3,\dots$) representing the order of the fringe. We can rearrange this to solve for the wavelength:

We are told that the light used falls within the visible spectrum, specifically between $380\text{ nm}$ and $740\text{ nm}$. Our job now is to test different integer values for $n$ to see which ones produce a wavelength in this valid range.

Let's hunt for the right integers:

If $n=3$, $\lambda =\frac{2500}{3}\approx 833\text{ nm}$. This is in the infrared region. Too large! If $n=4$, $\lambda =\frac{2500}{4}=625\text{ nm}$. This is a beautiful red-orange light. Perfect! If $n=5$, $\lambda =\frac{2500}{5}=500\text{ nm}$. This is a vibrant green light. Also perfect! If $n=6$, $\lambda =\frac{2500}{6}\approx 417\text{ nm}$. This is a deep violet light. Still visible! * If $n=7$, $\lambda =\frac{2500}{7}\approx 357\text{ nm}$. This drops into the ultraviolet region. Too small!

So, the possible visible wavelengths that can create a bright fringe at this exact angle are $625\text{ nm}$, $500\text{ nm}$, and $417\text{ nm}$.

Looking at our multiple-choice options, we need to find a pair that matches our findings. Option (d) offers the pair $625\text{ n-m}$ and $500\text{ n-m}$. This perfectly aligns with our $n=4$ and $n=5$ calculations.

Physics is incredibly elegant. At that exact angle of $\frac{1}{40}\text{ rad}$, the 4th order bright fringe of red light perfectly overlaps with the 5th order bright fringe of green light!