Question

In a Young's double slit experiment, the slits are separated by 0.2 mm and placed 2.0 m away. The distance between the central bright fringe and the first bright fringe (fringe width) is measured to be 1.5 cm. Determine the wavelength of light used in the experiment.

• A. 4200 Å
• B. 5000 Å
• C. 4600 Å
• D. 15000 Å (Calculated result)

Hint:

Always ensure your units are consistent. Convert mm and cm to meters (\textbf{m}) before plugging them into the formula to avoid power-of-ten errors.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept: 
In Young's Double Slit Experiment (YDSE), the fringe width ($\beta$) is the distance between two consecutive bright or dark fringes. 
Step 2: Formula Application: 
$\beta = \frac{\lambda D}{d} \implies \lambda = \frac{\beta d}{D}$ 
Step 3: Explanation: 
Given: $d = 0.2$ mm = $2 \times 10^{-4}$ m, $D = 2.0$ m, $\beta = 1.5$ cm = $1.5 \times 10^{-2}$ m. $$\lambda = \frac{(1.5 \times 10^{-2}) \times (2 \times 10^{-4})}{2.0}$$ $$\lambda = 1.5 \times 10^{-6} \text{ m}$$ Converting to Angstroms ($1 \text{ m} = 10^{10} \text{ \AA}$): $\lambda = 15000 \text{ \AA}$. 
Step 4: Final Answer: 
The wavelength is $15000 \text{\AA}$.