Question

In a Young’s double-slit experiment, a beam of light consisting of two wavelengths 500 nm and 600 nm is used. The interference fringes are observed at a screen placed 1.8 m away from the plane of slits (slit separation 0.3 mm). Calculate the least distance from the central maximum where the bright fringes due to both the wavelengths coincide.

Hint:

The coincidence formula $n_1 \lambda_1 = n_2 \lambda_2$ is extremely common in exams. Remember that the larger integer order always corresponds strictly to the shorter wavelength.

Answer 1

Concept:
• In Young's Double Slit Experiment, different wavelengths of incident light will simultaneously produce their own entirely independent interference patterns on the shared observation screen.

• Because the two distinct patterns possess drastically different intrinsic fringe widths, their respective bright maxima will mostly fall at entirely different spatial locations.

• However, at certain specific distances from the center, a bright fringe of one wavelength will perfectly overlap and physically coincide with a bright fringe of the second wavelength.

• This perfect coincidence mathematically occurs exactly when the spatial distance $y$ from the central maximum is absolutely identical for both independent interference patterns.

Step 1: List the provided experimental parameters strictly in SI units
First wavelength, $\lambda_1 = 500 \text{ nm} = 500 \times 10^{-9} \text{ m}$.
Second wavelength, $\lambda_2 = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$.
Screen distance, $D = 1.8 \text{ m}$.
Slit separation, $d = 0.3 \text{ mm} = 0.3 \times 10^{-3} \text{ m} = 3 \times 10^{-4} \text{ m}$.

Step 2: Establish the mathematical condition for coincidence
The generalized position for the $n$-th bright fringe measured directly from the central maximum is formally given by the established formula:
\[ y_n = n \frac{\lambda D}{d} \]
For perfect spatial coincidence to occur, the precise distance of the $n_1$-th bright fringe strictly of wavelength $\lambda_1$ must exactly equal the precise distance of the $n_2$-th bright fringe strictly of wavelength $\lambda_2$:
\[ y_{n_1} = y_{n_2} \]
\[ n_1 \frac{\lambda_1 D}{d} = n_2 \frac{\lambda_2 D}{d} \]

Step 3: Determine the lowest possible integer ratio
Since the physical parameters $D$ and $d$ are identical for both setups, they elegantly cancel out entirely from both sides of the equation:
\[ n_1 \lambda_1 = n_2 \lambda_2 \]
We gracefully rearrange this to find the strict ratio of the fringe orders:
\[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} \]
Substitute the given wavelength values to establish the numerical ratio:
\[ \frac{n_1}{n_2} = \frac{600 \text{ nm}}{500 \text{ nm}} = \frac{6}{5} \]
To rigorously find the *least* possible distance for the very first coincidence event, we absolutely must select the smallest possible integers that satisfy this exact fractional ratio. These are clearly:
\[ n_1 = 6 \text{ and } n_2 = 5 \]
This physically implies that the 6th bright fringe of the shorter 500 nm light perfectly overlaps the 5th bright fringe of the longer 600 nm light.

Step 4: Calculate the actual physical least distance
We can now calculate the explicit spatial distance $y_{min}$ using either $n_1$ or $n_2$. We will deliberately use $n_1$ for this calculation:
\[ y_{min} = n_1 \frac{\lambda_1 D}{d} \]
Carefully substitute all the rigorously converted SI values directly into the formula:
\[ y_{min} = 6 \times \frac{500 \times 10^{-9} \times 1.8}{0.3 \times 10^{-3}} \]
Let's aggressively simplify the numerical coefficients first:
\[ \frac{1.8}{0.3} = 6 \]
Now plug this factor squarely back into the calculation:
\[ y_{min} = 6 \times 500 \times 10^{-9} \times 6 \times 10^3 \]
\[ y_{min} = 36 \times 500 \times 10^{-6} \]
\[ y_{min} = 18000 \times 10^{-6} \text{ m} \]
For drastically better readability and to match standard conventions, convert this final result into millimeters:
\[ y_{min} = 18 \times 10^{-3} \text{ m} = 18 \text{ mm} \]

Step 5: Conclusion
The very first spatial location proceeding outwards from the central maximum where the two bright interference fringes will perfectly overlap is exactly 18 mm away.