Question

In a single-slit diffraction experiment, the diffraction pattern is observed on a screen placed at a distance of $2\text{ m}$ from a slit of width $1\text{ mm}$. If the distance between the first dark fringe on either side of the central bright fringe is $2\text{ mm}$, what is the wavelength of the monochromatic light used in this experiment?

• A. $5500\text{ }^\circ\text{A}$
• B. $11000\text{ }^\circ\text{A}$
• C. $1100\text{ }^\circ\text{A}$
• D. $3900\text{ }^\circ\text{A}$

Hint:

Always read fringe problem descriptions carefully! The distance "between the first dark minima on either side" is exactly double the individual step width, which matches the definition of the central maximum width.

Answer 1

The Correct Option is A

Understanding the Concept: In single-slit diffraction, the position of the first minimum (dark fringe) relative to the center is given by $y_1 = \frac{\lambda D}{a}$. The total width of the central bright maximum is defined as the distance stretching from the first dark fringe on one side to the first dark fringe on the opposite side: \[ W = 2y_1 = \frac{2\lambda D}{a} \]
Step 1: Isolate the target variable $\lambda$ and substitute matching SI metrics.
We are given:
Screen distance, $D = 2\text{ m}$
Slit width, $a = 1\text{ mm} = 1 \times 10^{-3}\text{ m}$
Total central maximum width, $W = 2.2\text{ mm} = 2.2 \times 10^{-3}\text{ m}$
Rearranging the formula for $\lambda$: \[ \lambda = \frac{W \cdot a}{2D} \]
Step 2: Perform calculation evaluation.
\[ \lambda = \frac{(2.2 \times 10^{-3}) \times (1 \times 10^{-3})}{2 \times 2} = \frac{2.2 \times 10^{-6}}{4} = 0.55 \times 10^{-6}\text{ m} \] \[ \lambda = 550 \times 10^{-9}\text{ m} = 5500 \times 10^{-10}\text{ m} = 5500\text{ }^\circ\text{A} \]