Question

A lens \((n=1.52)\) is coated with a magnesium fluoride film \((n=1.38) .\) (a) If the coating is to cause destructive interference in reflected light for \(\lambda=560 \mathrm{nm}\) (the peak of the solar spectrum), what should its minimum thickness be? (b) At what two wavelengths closest to 560 nm does the coating cause constructive interference in reflected light? (c) Is any visible light reflected? Explain.

Short answer

Using the formula for destructive interference, we plug in the values: \(t = (\frac{1}{2}) \frac{560 \text{ nm}}{2(1.38 - 1)}\) \(t = \frac{1}{2} \frac{560 \text{ nm}}{0.76}\) \(t = 367.1 \text{ nm}\) So, the minimum thickness for destructive interference is 367.1 nm. (b) Find two closest wavelengths for constructive interference Using the formula for constructive interference and the value of t = 367.1 nm, we choose two closest integer values of m, one smaller (m=0), and one larger (m=1) than the value used in part (a): For m=0: \(\lambda = \frac{2(367.1 \text{ nm})(1.38 - 1)}{0}\) Since we can't divide by zero, there is no wavelength for m=0. For m=1: \(\lambda = \frac{2(367.1 \text{ nm})(1.38 - 1)}{1}\) \(\lambda = 758.6 \text{ nm}\) So, the two closest wavelengths for constructive interference are not possible for m=0, and 758.6 nm for m=1. (c) Is any visible light reflected? The visible light spectrum ranges from about 400 nm (violet) to 700 nm (red). Our constructive interference wavelength is 758.6 nm, which is outside the range of visible light. Therefore, no visible light is reflected.

Step-by-step solution

(a) Find minimum thickness for destructive interference

For a thin film, destructive interference occurs when the optical path difference between the two reflected rays is an odd multiple of λ/2. Using the formula for destructive interference: \(t = (m + \frac{1}{2}) \frac{\lambda}{2(n_{film} - n_{air})}\) Where t is the thickness of the film, m is an integer (for minimum thickness, we choose the smallest integer, m = 0), λ is the wavelength of light (λ = 560 nm), n_film = 1.38 (refractive index of the film), and n_air = 1 (refractive index of air). Now, we plug in the values: \(t = (\frac{1}{2}) \frac{560 \text{ nm}}{2(1.38 - 1)}\) Calculate the thickness t.

(b) Find two closest wavelengths for constructive interference

For a thin film, constructive interference occurs when the optical path difference between the two reflected rays is an integer multiple of λ. Using the formula for constructive interference: \(t = m \frac{\lambda}{2(n_{film} - n_{air})}\) We can rearrange the formula to solve for λ: \(\lambda = \frac{2t(n_{film} - n_{air})}{m}\) Using the t obtained in part (a), we can find two wavelengths on either side of 560 nm that cause constructive interference. Choose two closest integer values of m, one smaller and one larger than the value used in part (a). Calculate the two wavelengths corresponding to these m values.

(c) Determine if any visible light is reflected

The visible light spectrum ranges from about 400 nm (violet) to 700 nm (red). Compare the wavelengths obtained in part (b) with this range. If any of the wavelengths fall within the visible light spectrum, then visible light is reflected. Otherwise, no visible light is reflected. Explain your conclusion based on the calculated wavelengths.