Question

Two coherent sources \(O_1\) and \(O_2\) in Young's double slit experiment are illuminated with monochromatic light of wavelength \(5000\ \text{\AA}\). If a second order dark fringe is formed at a point \(R\) on the screen, the path difference \[ O_1R \sim O_2R \] is

• A. \(7.5\ \mu\text{m}\)
• B. \(0.75\ \mu\text{m}\)
• C. \(0.075\ \mu\text{m}\)
• D. \(75\ \mu\text{m}\)

Hint:

In YDSE: \[ \text{Bright fringe: } \Delta=n\lambda \] \[ \text{Dark fringe: } \Delta=(2n-1)\frac{\lambda}{2} \] Always substitute the correct order carefully.

Answer 1

The Correct Option is B

Step 1: Recall the path difference condition for dark fringes in YDSE.
Destructive interference occurs when path difference is an odd half-wavelength: $\Delta = (2n-1)\frac{\lambda}{2}$, $n = 1, 2, 3, \ldots$
Step 2: Identify the order of the dark fringe.
Second-order dark fringe: $n = 2$.
Step 3: Compute the path difference.
\[ \Delta = (2 \times 2 - 1)\frac{\lambda}{2} = \frac{3\lambda}{2} \]
Step 4: Convert wavelength to SI units.
$\lambda = 5000$ A $= 5 \times 10^{-7}$ m.
Step 5: Calculate the path difference in micrometres.
\[ \Delta = \frac{3 \times 5 \times 10^{-7}}{2} = 7.5 \times 10^{-7} \text{ m} = 0.75 \text{ } \mu\text{m} \]
Step 6: State the final answer.
\[ \boxed{\Delta = 0.75 \ \mu\text{m}} \]