Question

In a double-slit experiment, He-Ne laser light of wavelength \(633 \mathrm{nm}\) produced an interference pattern on a screen placed at some distance from the slits. When one of the slits was covered with a thin glass slide of thickness \(12.0 \mu \mathrm{m},\) the central fringe shifted to the point occupied earlier by the 10 th dark fringe (see figure). What is the refractive index of the glass slide? (a) Without the glass slide (b) With glass slide

Short answer

Answer: The refractive index of the glass slide is approximately 1.53.

Step-by-step solution

Identifying known values and the formula for path difference

We are given: - Wavelength of light (λ) = 633 nm = 633 x 10^-9 m - Thickness of the glass slide (t) = 12.0 μm = 12.0 x 10^-6 m - The central fringe shifts to the previous 10th dark fringe when the glass slide is placed. We need to find the path difference (Δ) caused by the glass slide. The formula for path difference in a double-slit experiment is given by: Δ = mλ where m = the order of the dark fringe (in this case, m=10).

Calculate the path difference

Using the given values and the path difference formula, we can calculate the path difference: Δ = 10λ = 10 * (633 * 10^-9 m) = 6.33 * 10^-6 m

Determine the extra distance traveled in the glass slide

When the glass slide is placed over one slit, the light from that slit has to travel an extra distance (d) through the slide compared to the other slit. This extra distance is given by: d = (n - 1) t where n = refractive index of the glass slide.

Use the path difference to find the refractive index

We have the path difference (Δ) and the extra distance (d) formulas: Δ = 6.33 * 10^-6 m d = (n - 1) t Since the path difference is equal to the extra distance traveled in the glass slide, we can equate the two formulas: 6.33 * 10^-6 m = (n - 1) * (12.0 * 10^-6 m) Now, we can solve for the refractive index (n): n = (6.33 * 10^-6 m / (12.0 * 10^-6 m)) + 1

Calculate the refractive index

Plugging the values into the equation, we get: n = (6.33 / 12) + 1 = 1.5275 Thus, the refractive index of the glass slide is approximately 1.53.

Key concepts

Interference Pattern

The double-slit experiment is famous for demonstrating the wave nature of light through an interference pattern. When light passes through two closely spaced slits, it divides into two beams that overlap on a screen placed at some distance. This overlapping of light waves leads to constructive and destructive interference.

- **Constructive interference** occurs when the crests of one wave line up with the crests of another, creating bright fringes or bands.
- **Destructive interference** happens when the crests of one wave align with the troughs of another, resulting in dark fringes or bands.

The resulting pattern of alternating bright and dark fringes is known as the interference pattern. This phenomenon vividly showcases how waves interact with each other. It's important to note that the pattern's formation depends significantly on the light's wavelength and the distance between the slits.

Refractive Index

The refractive index is a measure of how much light bends, or refracts, when entering a material. Each material has a specific refractive index that describes how much slower light travels through it compared to in a vacuum.

- For the glass slide in the problem, the refractive index influences how the light waves behave when passing through.
- Mathematically, the refractive index ( ) can be calculated if you know the path difference caused by passing through the material.

In the given exercise, covering one slit with the glass slide caused a shift in the interference pattern. Calculating the refractive index allows us to understand how that glass slide alters the light path compared to air. Understanding this concept is crucial as it has applications ranging from designing lenses to creating optical instruments.

Path Difference

Path difference is a key concept in understanding interference patterns. It refers to the difference in the distance traveled by two light waves coming from the slits before they meet and overlap on a screen.

- When waves have different path lengths, they interfere differently. - Constructive interference happens when the path difference is an integer multiple of the wavelength, while destructive interference occurs when there is a half-integer multiple.

In the problem, when a glass slide is added to one of the slits, it causes light traveling through that slit to experience a different path length. The path difference created here is vital in determining how much the interference pattern shifts. In this case, it's exactly matching the previous position of the 10th dark fringe.

Wavelength

Wavelength is the distance between consecutive crests of a wave, and it's fundamental to the behavior of light. It's a critical component in calculating and understanding interference patterns, as well as the overall behavior of light in wave optics.

- In the given problem, the wavelength of the light used is 633 nm.
- Wavelength influences both the spacing and position of the fringes in an interference pattern.

Light of different wavelengths will produce fringes differently spaced apart. Adjusting the wavelength, for instance, changes where the constructive and destructive interferences occur. Thus, knowing the wavelength helps predict and explain the shifts seen in the interference pattern when additional layers like glass are introduced.