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

The problem involves finding the position where the minima (dark fringes) corresponding to two different wavelengths coincide in Young’s Double Slit Experiment (YDSE).

The condition for minimum intensity for a given wavelength \(\lambda\) is:

\(d \sin\theta = \left(m + \dfrac{1}{2}\right)\lambda\)

where \(d\) is the slit separation, \(\theta\) is the angular position of the minima, and \(m\) is the order of the minimum.

For small angles, \(\sin\theta \approx \tan\theta = \dfrac{x}{L}\), where \(x\) is the distance from the central maximum and \(L\) is the distance between the slits and the screen.

Hence, the condition for minima becomes:

\(d \dfrac{x}{L} = \left(m + \dfrac{1}{2}\right)\lambda\)

For the minima of both wavelengths to coincide, their positions must be equal:

\(\lambda_1\left(m_1 + \dfrac{1}{2}\right) = \lambda_2\left(m_2 + \dfrac{1}{2}\right)\)

Given:

• \(\lambda_1 = 450\,\text{nm}\)
• \(\lambda_2 = 550\,\text{nm}\)

\(450(m_1 + \tfrac{1}{2}) = 550(m_2 + \tfrac{1}{2})\)

Assuming \(m_1 = m_2 = m\):

\(450m + 225 = 550m + 275\)

\(m = 0.5\)

The position of the coincident minima is given by:

\(x = \dfrac{m + \tfrac{1}{2}}{L} \cdot \dfrac{\lambda_1 \lambda_2}{\lambda_2 - \lambda_1}\)

Substituting the values:

\(x = \dfrac{0.5 + 0.5}{1.5} \cdot \dfrac{450 \times 10^{-9} \times 550 \times 10^{-9}} {(550 - 450)\times 10^{-9}}\)

\(x \approx 1.65\,\text{mm}\)

Therefore, the distance from the central maximum where the minima of both wavelengths coincide is \(1.65\,\text{mm}\).