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.

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.