# a) A film of magnesium fluoride, (n=1.38), 1.25 x 10^{-5} cm thick, is used to coat a camera lens...

## Question:

a) A film of magnesium fluoride, (n=1.38), 1.25 x {eq}10^{-5} {/eq} cm thick, is used to coat a camera lens (n=1.55).

Are any wavelengths in the visible spectrum intensified in the reflected light?

b) Bats use the reflections from ultra-high frequency sound to locate their prey.

Estimate the typical frequency of a bat's sonar. Take the speed of sound to be 3.40 x {eq}10^2 {/eq} m/s, and a small moth 3.00 mm across, to be a typical target.

## Interference of Waves:

Two waves of the same frequency and that are coherent will form a stationary pattern of maxima and minima that correspond to two waves being in phase or out of phase. This phenomenon is called an interference of waves.

## Answer and Explanation:

a) The condition for the light to interfere constructively upon reflection can be written as follows:

{eq}2 d n = m \lambda {/eq}

Here

- {eq}d = 1.25\times 10^{-7} \ m {/eq} is the thickness of the film;

- {eq}n = 1.38 {/eq} is the index of refraction of the coating film;

The wavelengths that will be intensified in reflection are given by:

{eq}\lambda = \dfrac {2dn}{m} {/eq}

Substituting the numbers, we get:

{eq}\lambda = \dfrac {2\times 1.25\times 10^{-7} \ m \times 1.38}{m} = \dfrac {3.45 \times 10^{-7} \ m}{m} {/eq}

It is clear that the largest wavelength intensified in reflection is

{eq}\lambda_{max} = \dfrac {3.45\times 10^{-7} \ m}{1} = 3.45\times 10^{-7} \ m {/eq}

This wavelength corresponds to the ultraviolet light, so **no wavelengths in the visible spectrum will be intensified in reflection**.

b) The sound wave will effectively reflect from the target if the target is larger than the wavelength of the wave. Estimate for the frequency of a bat's sound wave can be made as follows:

{eq}f = \dfrac {c}{\lambda} = \dfrac {c}{d} {/eq}

Here

- {eq}c = 340 \ m/s {/eq} is the speed of sound in the air;

- {eq}d = 3 \ mm {/eq} is the size of the prey for the bat;

Evaluating, we get:

{eq}f = \dfrac {340 \ m/s}{3\times 10^{-3} \ m} \approx \boxed{1.13\times 10^5 \ Hz} {/eq}

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from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 8 / Lesson 16