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Chapter 25: Problem 18

A transparent film \((n=1.3)\) is deposited on a glass lens \((n=1.5)\) to form a nonreflective coating. What is the minimum thickness that would minimize reflection of light with wavelength \(500.0 \mathrm{nm}\) in air?

Short Answer

Expert verified
Answer: The minimum thickness of the nonreflective coating is approximately 96.15 nm.

Step by step solution

01

Calculate the wavelength of light in the thin film

The refractive index of the film is given by \(n_{film} = 1.3\). The wavelength of light in air is given by \(\lambda_{air} = 500 \thinspace nm\). The wavelength of light in the thin film is given by: $$ \lambda_{film} = \frac{\lambda_{air}}{n_{film}}. $$ Plug in the given values and solve for \(\lambda_{film}\): $$ \lambda_{film} = \frac{500 \thinspace nm}{1.3}. $$
02

Calculate the minimum thickness of the thin film

To minimize reflection, the path difference between the light waves reflected from both the surfaces of the thin film must be equal to half of the wavelength in the thin film, or \(\frac{\lambda_{film}}{2}\). Since the light travels through the thin film twice before exiting (once going in and once coming out), the path difference is equal to \(2t\) (where \(t\) is the thickness of the film). Therefore, we have: $$ 2t = \frac{\lambda_{film}}{2}. $$ Divide by 2 and plug in the value of \(\lambda_{film}\) from Step 1: $$ t = \frac{\frac{500 \thinspace nm}{1.3}}{4}. $$
03

Calculate the result

Finally, perform the calculations to find the minimum thickness \(t\): $$ t = \frac{500 \thinspace nm}{1.3 \times 4} = \frac{500}{5.2} \thinspace nm \approx 96.15 \thinspace nm. $$ The minimum thickness that would minimize reflection of light with wavelength \(500.0 \thinspace nm\) in air is about \(96.15 \thinspace nm\).

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