Question

Two light sources of wavelengths \(450\,\text{nm}\) and \(550\,\text{nm}\) are used for YDSE with slit separation \(2.25\,\text{mm}\) and distance between the slits and the screen is \(1.5\,\text{m}\). Then the distance from central maxima for which minima of both wavelengths coincide is:

• A. \(1.65\,\text{mm}\)
• B. \(1.55\,\text{mm}\)
• C. \(1.45\,\text{mm}\)
• D. \(1.85\,\text{mm}\)

Hint:

For coincidence of minima of two wavelengths in YDSE, equate \((2m+1)\lambda\) for both waves and use the smallest integer solution.

Answer 1

The Correct Option is A

Concept:
In Young’s Double Slit Experiment, the position of the \(n^{\text{th}}\) minima is given by: \[ y = \left(m + \frac{1}{2}\right)\frac{\lambda D}{d} \] For minima of two different wavelengths to coincide, their positions must be equal.
Step 1: Condition for Coincidence of Minima
For wavelengths \(\lambda_1 = 450\,\text{nm}\) and \(\lambda_2 = 550\,\text{nm}\): \[ \left(m_1 + \frac{1}{2}\right)\lambda_1 = \left(m_2 + \frac{1}{2}\right)\lambda_2 \] \[ (2m_1 + 1)\lambda_1 = (2m_2 + 1)\lambda_2 \] Substitute values: \[ (2m_1 + 1)\times 450 = (2m_2 + 1)\times 550 \] Divide by 50: \[ (2m_1 + 1)\times 9 = (2m_2 + 1)\times 11 \] Smallest integer solution: \[ 2m_1 + 1 = 11 \Rightarrow m_1 = 5 \] \[ 2m_2 + 1 = 9 \Rightarrow m_2 = 4 \]
Step 2: Calculate Distance from Central Maxima
Using wavelength \(450\,\text{nm}\): \[ y = \left(5 + \frac{1}{2}\right)\frac{450 \times 10^{-9} \times 1.5}{2.25 \times 10^{-3}} \] \[ y = \frac{5.5 \times 450 \times 1.5}{2.25}\times 10^{-6} \] \[ y = 1.65 \times 10^{-3}\,\text{m} = 1.65\,\text{mm} \] \[ \boxed{y = 1.65\,\text{mm}} \]