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: Write the condition for dark fringes in YDSE.
In Young's double slit experiment, the path difference for dark fringes is \[ \Delta=(2n-1)\frac{\lambda}{2} \] where \[ n=1,2,3,\dots \] For the second order dark fringe, \[ n=2 \] Thus, \[ \Delta=(2\times 2-1)\frac{\lambda}{2} \] \[ \Delta=\frac{3\lambda}{2} \]

Step 2: Convert wavelength into SI units.
Given wavelength: \[ \lambda=5000\ \text{\AA} \] We know that \[ 1\ \text{\AA}=10^{-10}\ \text{m} \] Hence, \[ \lambda=5000\times 10^{-10}\ \text{m} \] \[ \lambda=5\times 10^{-7}\ \text{m} \]

Step 3: Calculate the path difference.
Substituting the value of \(\lambda\), \[ \Delta=\frac{3}{2}\times 5\times 10^{-7} \] \[ \Delta=7.5\times 10^{-7}\ \text{m} \] Now, \[ 1\ \mu\text{m}=10^{-6}\ \text{m} \] Therefore, \[ \Delta=0.75\ \mu\text{m} \]

Step 4: Final conclusion.
Hence, the required path difference is \[ \boxed{0.75\ \mu\text{m}} \]