AIIMS 2003 Physics Interference MCQ Question

In a double slit experiment, the interference pattern is created by the superposition of light waves emanating from two closely spaced slits. The position of the interference maxima is influenced by the path difference between the two beams. When a thin film with a different refractive index is introduced in the path of one of the beams, it alters the effective optical path length, leading to a shift in the interference pattern.

Explanation of the Correct Answer (C)

1. Understanding Optical Path Length: When light passes through a medium of refractive index $n$, the speed of light decreases, which affects the optical path length. The optical path length $L$ in a medium is given by: $L = n \cdot d$ where $d$ is the physical thickness of the medium.

2. Calculating the Optical Path Length in the Thin Film: For the thin film introduced in the upper beam:

   • Thickness $d = 2 \, \mu m = 2 \times 10^{-6} \, m$
   • Refractive index $n = 1.5$

   The optical path length through the thin film is: $L_{\text{film}} = n \cdot d = 1.5 \cdot (2 \times 10^{-6}) = 3 \times 10^{-6} \, m$

3. Calculating the Additional Optical Path Introduced: When light travels through the thin film, the additional path length compared to traveling in air (where $n \approx 1$) is: $\Delta L = L_{\text{film}} - d = (3 \times 10^{-6}) - (2 \times 10^{-6}) = 1 \times 10^{-6} \, m$

4. Calculating Phase Change: The additional path length leads to a phase change. The phase difference $\Delta \phi$ introduced is given by: $\Delta \phi = \frac{2\pi}{\lambda} \Delta L$ where $\lambda = 500 \, nm = 500 \times 10^{-9} \, m$.

   Substituting the values: $\Delta \phi = \frac{2\pi}{500 \times 10^{-9}} \cdot (1 \times 10^{-6})$ $\Delta \phi = \frac{2\pi}{500} \cdot 10^{-3} \approx 0.01257 \, \text{radians}$

5. Impact on Interference Pattern: Each complete shift of $2\pi$ represents one fringe shift. To find the number of fringes shifted: $\text{Fringe shift} = \frac{\Delta \phi}{2\pi} = \frac{0.01257}{2\pi} \approx 0.002 \, \text{fringes}$ Since the question involves a total optical path change due to the film, we also account for the fact that the upper beam experiences a half-wavelength phase change upon reflection at the film interface (air to a denser medium): $\Delta L_{\text{effective}} = \Delta L + \frac{\lambda}{2} = (1 \times 10^{-6}) + (250 \times 10^{-9}) = 1.25 \times 10^{-6} \, m$

   The corresponding fringe shift would be approximately 2 fringes downward due to this phase alteration.

Clarification of Incorrect Options

• Option A: Remain unshifted: This option is incorrect because the introduction of the thin film alters the optical path length, which directly affects the interference pattern.

• Option B: Shift downward by nearly two fringes: This option is misleading; while the shift does occur, it should be noted that the effective shift is upward by nearly two fringes due to the phase change caused by reflection.

• Option D: Shift downward by 10 fringes: This option is also incorrect. The calculations show that the shift is not significant enough to cause a 10-fringe displacement; the effect of the thin film results in a shift that is much smaller.

Conclusion

Thus, the correct answer is C) shift upward by nearly two fringes, as the introduction of the thin film introduces a phase change that alters the position of the interference maximum.