Question

Thin Films At a science museum, Marlow looks down into a display case and sees two pieces of very flat glass lying on top of each other with light and dark regions on the glass. The exhibit states that monochromatic light with a wavelength of \(550 \mathrm{nm}\) is incident on the glass plates and that the plates are sitting in air. The glass has an index of refraction of \(1.51 .\) (a) What is the minimum distance between the two glass plates for one of the dark regions? (b) What is the minimum distance between the two glass plates for one of the light regions? (c) What is the next largest distance between the plates for a dark region? [Hint: Do not worry about the thickness of the glass plates; the thin film is the air between the plates.]

Short answer

Answer: The minimum distance between the glass plates for a dark region is 137.5 nm, for a light region is 275 nm, and the next largest distance for a dark region is 412.5 nm.

Step-by-step solution

(a) Step 1: Finding the minimum distance for a dark region

First, let's consider the minimum distance for destructive interference, which creates a dark region. Destructive interference occurs when the path difference between the light rays is an odd multiple of half the wavelength, i.e., it equals \((2n-1)\frac{\lambda}{2}\), where \(n=1,2,3...\). For the minimum distance, we need to consider the first dark region, i.e., \(n=1\). The path difference is then \((2\cdot 1-1)\frac{550 \mathrm{nm}}{2} = 550 \mathrm{nm}/2\).

(a) Step 2: Converting the path difference to distance

Now, we have the path difference. We can relate it to the distance between the glass plates as follows: \(2d = (2n-1)\frac{\lambda}{2}\), where \(d\) is the distance between the plates, and the factor of 2 accounts for the round trip the light makes between the plates. In this case, \(2d = 550 \mathrm{nm}/2\). Solving for \(d\), we find that the minimum distance between the glass plates for a dark region is \(d = 550 \mathrm{nm}/(2\cdot 2) = \boxed{137.5 \mathrm{nm}}\).

(b) Step 3: Finding the minimum distance for a light region

Now, let's consider the minimum distance for constructive interference, which creates a light region. Constructive interference occurs when the path difference between the light rays is an even multiple of half the wavelength, i.e., it equals \(n\frac{\lambda}{2}\), where \(n=1,2,3...\). For the minimum distance, we need to consider the first light region, i.e., \(n=1\). The path difference is then \(1\cdot\frac{550 \mathrm{nm}}{2} = 550 \mathrm{nm}/2\).

(b) Step 4: Converting the path difference to distance

Applying the same relationship between path difference and distance as in Step 2, we get: \(2d = n\frac{\lambda}{2}\). In this case, \(2d = 550 \mathrm{nm}/2\). Solving for \(d\), we find that the minimum distance between the glass plates for a light region is \(d = 550 \mathrm{nm}/(2\cdot 1) = \boxed{275 \mathrm{nm}}\).

(c) Step 5: Finding the next largest distance for a dark region

For the next largest distance for a dark region, we need to consider the case with \(n=2\). The path difference in this case is given by \((2\cdot 2-1)\frac{550 \mathrm{nm}}{2} = 3\cdot 550 \mathrm{nm}/2\).

(c) Step 6: Converting the path difference to distance

Applying the same relationship between path difference and distance as in Steps 2 and 4, we get: \(2d = (2n-1)\frac{\lambda}{2}\). In this case, \(2d = 3\cdot 550 \mathrm{nm}/2\). Solving for \(d\), we find that the next largest distance between the glass plates for a dark region is \(d = 3\cdot 550 \mathrm{nm}/(2\cdot 2) = \boxed{412.5 \mathrm{nm}}\).