In Young's double slit experiment, when a thin transparent sheet is placed in front of the slits, it alters the path of light waves passing through, leading to a shift in the interference pattern. Let's break down how to find the thickness of this sheet based on the given information.
Understanding the Shift in the Interference Pattern
The interference pattern consists of bright and dark fringes, which are formed due to constructive and destructive interference of light waves. The central fringe is typically the brightest spot, and it corresponds to the point where the path difference between the two waves is zero.
When you introduce a thin sheet with a refractive index (n) in front of one of the slits, it changes the effective wavelength of light passing through that slit, which in turn shifts the position of the fringes. The formula to calculate the new position of the fringes due to this shift can be expressed as:
Key Formula for Path Difference
The path difference (Δ) introduced by a sheet of thickness (t) and refractive index (n) is given by:
This path difference causes a shift in the fringe pattern. Each fringe shift corresponds to a change in path difference of one wavelength (λ). Therefore, if the central fringe shifts to the position of the fifth dark fringe, we can relate this to the number of wavelengths by:
Relating Path Difference to Fringe Shift
The condition for dark fringes in a double-slit experiment is given by:
- Path difference = (m + 1/2) * λ, where m is the order of the dark fringe.
For the fifth dark fringe (m = 4), the path difference would be:
- Δ = (4 + 1/2) * λ
- Δ = 4.5 * λ
Calculating the Thickness of the Sheet
Now, substituting the values we have into our equations, we know that the path difference introduced by the sheet must equal the path difference corresponding to the fifth dark fringe:
Given that the wavelength (λ) is 589 nm (which is 589 x 10-9 meters) and the refractive index (n) is 1.5, we can plug these values into the equation:
- (1.5 - 1) * t = 4.5 * (589 x 10-9)
This simplifies to:
- 0.5 * t = 4.5 * (589 x 10-9)
- t = (4.5 * 589 x 10-9) / 0.5
Now calculate the thickness:
- t = (4.5 * 589 x 10-9) / 0.5 = 4.5 * 1178 x 10-9
- t = 5301 x 10-9 m = 5.301 x 10-6 m = 5.301 μm
Final Result
The thickness of the thin transparent sheet is approximately 5.301 micrometers. This process shows how the introduction of a medium with a different refractive index can significantly impact the behavior of light in an interference experiment, leading to interesting and measurable shifts in the resulting patterns.
