Question

In Young's double slit experiment, monochromatic light of 500 nm falls on the slits separated by a distance of 2 mm. If the screen is 2 m away from the source, then the fringe width is

• A. 5 mm
• B. 2.0 mm
• C. 0.5 mm
• D. 2.5 mm
• E. 1.5 mm

Hint:

Increasing the distance to the screen ($D$) increases the fringe width.

Answer 1

The Correct Option is C

Step 1: Concept
The fringe width ($\beta$) in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$.

Step 2: Meaning
$\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m}$, $D = 2 \text{ m}$, and $d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$.

Step 3: Analysis
$\beta = \frac{500 \times 10^{-9} \times 2}{2 \times 10^{-3}} = 500 \times 10^{-6} \text{ m} = 0.5 \times 10^{-3} \text{ m} = 0.5 \text{ mm}$.

Step 4: Conclusion
Hence, the fringe width is 0.5 mm.
Final Answer: (C)