Question

In a double-slit experiment, He-Ne laser light of wavelength \(633 \mathrm{nm}\) produced an interference pattern on a screen placed 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 bright fringe shifted to the point occupied earlier by the 10 th dark fringe (see the figure). What is the index of refraction of the glass slide? (a) Without the glass slide (b) With glass slide

Short answer

Answer: The index of refraction of the glass slide is approximately 1.49.

Step-by-step solution

Understand the basic concepts of the double-slit experiment

In the double-slit experiment, light passes through two narrow slits and interferes, creating a pattern of bright and dark fringes on a screen. The condition for a bright fringe is given by: \(\frac{dsin\theta}{m\lambda} = 1\), where \(d\) is the distance between the slits, \(\theta\) is the angle of the fringe from the central maximum, \(m\) is the order of the fringe, and \(\lambda\) is the wavelength of the light.

Analyze the fringe shift effect with the glass slide

When the glass slide is inserted over one of the slits, it shifts the interference pattern. The shift is caused by a change in the path difference between the two slits due to the glass slide's refractive index, which slows down the light and increases the optical path length. Let's denote the refractive index of the glass slide as \(n\) and its thickness as \(t\). The change in the optical path length can be represented as \((n-1)t\). This shift will cause existing fringes on the screen to move to new positions.

Calculate the fringe shift

The central bright fringe shifts to the position occupied by the 10th dark fringe. For a dark fringe, the condition is given by: \(\frac{dsin\theta}{(m+\frac{1}{2})\lambda} = 1\). In this case, the central bright fringe corresponds to \(m = 0\), and the 10th dark fringe corresponds to \(m = 9\). Therefore, the change in path difference due to the shift is equal to the path difference between these two fringes, which is given by \(\Delta(m+\frac{1}{2})\lambda\). In this case, \(\Delta{m}=9\), so the change in path difference is \((9+\frac{1}{2})\lambda\).

Use the fringe shift to calculate the refractive index

The change in path difference due to the fringe shift should be equal to the change in the optical path length caused by the glass slide. Therefore, we have \((n-1)t = (9+\frac{1}{2})\lambda\). We can rearrange this equation to solve for the refractive index \(n\). So, \(n = 1 + \frac{(9+\frac{1}{2})\lambda}{t}\). Now, we can plug in the given values for \(\lambda = 633 \mathrm{nm} = 6.33 \times 10^{-7} \mathrm{m}\) and \(t = 12.0 \mu \mathrm{m} = 12 \times 10^{-6} \mathrm{m}\).

Calculate the index of refraction

To find the index of refraction of the glass slide, plug in the given values into the equation in Step 4: \(n = 1 + \frac{(9+\frac{1}{2})(6.33 \times 10^{-7} \mathrm{m})}{12 \times 10^{-6} \mathrm{m}}\). Solving this equation gives us: \(n \approx 1.49\). Thus, the index of refraction of the glass slide is approximately \(1.49\).