Q. 66 Engineers are testing a new thin... [FREE SOLUTION]

Chapter 17: Q. 66 (page 486)

Engineers are testing a new thin-film coating whose index of refraction is less than that of glass. They deposit a 560-nm-thick layer on glass, then shine lasers on it. A red laser with a wavelength of 640 nm has no reflection at all, but a violet laser with a wavelength of 400 nm has a maximum reflection. How the coating behaves at other wavelengths is unknown. What is the coating鈥檚 index of refraction?

Short Answer

Expert verified

The coating's index of refraction is n= 1.429

Step by step solution

The concept of refraction

The change in direction of a wave passing from one medium to another or from a gradual change in the medium is known as the refraction.

Explanation of the solution

Here, $d = 560 n m$

$位_{r e d} = 640 n m$

$位_{b l u e} = 400 n m$

Thus, the solution of the refraction is $n = \frac{位_{e} m_{e}}{2 d}$

Thus, the ratio of the constructive interference is $\frac{位_{C}}{位_{D}} = \frac{\frac{\frac{2 n d}{m c}}{2 n d}}{m_{d} - \frac{1}{2}} \frac{400}{640} - \frac{m_{d} - \frac{1}{2}}{m c} \frac{m_{d} - \frac{1}{2}}{m c} = 0.625$

In order to find out the substitute of the $m_{C}$ by the random numbers of role="math" localid="1649153848269" $m_{D}$ . Thus, the substitute of $m_{D}$ is 3.

Thus, role="math" localid="1649153919628" $\frac{3 - \frac{1}{2}}{m c} = 1.6 m c = 4$