Question

In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890 Å is :

• A. 1178 × 10⁻¹² m
• B. 5890 × 10⁻⁷ m
• C. 1178 × 10⁻⁹ m
• D. 1178 × 10⁻⁶ m

Hint:

Distance between $n^{th}$ and $m^{th}$ bright fringes is $|n-m|\beta$. Always convert units to SI (meters) before calculating.

Answer 1

The Correct Option is D

Step 1: Distance between $n^{th}$ and $m^{th}$ bright fringe is $(n-m)\beta$, where $\beta = \frac{\lambda D}{d}$.
Step 2: Distance $= (3-1) \frac{\lambda D}{d} = \frac{2 \lambda D}{d}$.
Step 3: $\text{Dist} = \frac{2 \times 5890 \times 10^{-10} \times 0.5}{0.5 \times 10^{-3}} = 2 \times 5890 \times 10^{-7} = 11780 \times 10^{-7} = 1178 \times 10^{-6} \text{ m}$.