Question

When monochromatic light of wavelength \(600\;nm\) is used in Young's double slit experiment, the fifth order bright fringe is formed at \(6\;mm\) from the central bright fringe on the screen. If the experiment is conducted with light of wavelength \(400\;nm\), then the third order bright fringe will be located at

• A. \(1.6\;mm\)
• B. \(2\;mm\)
• C. \(2.4\;mm\)
• D. \(3\;mm\)

Hint:

In Young's double slit experiment, \[ y_n=n\beta \] and fringe width \(\beta\) is directly proportional to wavelength.

Answer 1

The Correct Option is C

Step 1: Recall the position of a bright fringe.
In Young's double slit experiment, the $n^{\text{th}}$ bright fringe lies at \[ y_n = \frac{n \lambda D}{d} \] where $\dfrac{\lambda D}{d}$ is the fringe width.
Step 2: See what stays fixed.
The slit separation $d$ and the slit-to-screen distance $D$ do not change between the two parts of the experiment, so $y_n \propto n \lambda$.
Step 3: Write the ratio of the two situations.
Comparing the new third-order fringe with the original fifth-order fringe, \[ \frac{y_{\text{new}}}{y_{\text{old}}} = \frac{n_2 \lambda_2}{n_1 \lambda_1} \]
Step 4: Insert the known values.
Here $n_1 = 5$, $\lambda_1 = 600\ \text{nm}$, $y_{\text{old}} = 6\ \text{mm}$, and $n_2 = 3$, $\lambda_2 = 400\ \text{nm}$, so \[ \frac{y_{\text{new}}}{6} = \frac{3 \times 400}{5 \times 600} \]
Step 5: Simplify the fraction.
The right side becomes $\dfrac{1200}{3000} = \dfrac{2}{5}$, giving \[ y_{\text{new}} = 6 \times \frac{2}{5} \]
Step 6: Compute the final position.
Therefore \[ y_{\text{new}} = 2.4\ \text{mm} \] and the third-order bright fringe is located at \[ \boxed{2.4\ \text{mm}} \]