Question

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’s index of refraction?

Short answer

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, $d=560nm$

${\lambda }_{red}=640nm$

${\lambda }_{blue}=400nm$

Thus, the solution of the refraction is $n=\frac{{\lambda }_e{m}_e}{2d}$

Thus, the ratio of the constructive interference is $$\begin{gathered} \frac{{\lambda }_C}{{\lambda }_D}=\frac{\frac{\frac{2nd}{mc}}{2nd}}{m_d-{\displaystyle \frac{1}{2}}} \\ \frac{400}{640}-\frac{m_d-\frac{1}{2}}{mc} \\ \frac{m_d-\frac{1}{2}}{mc}=0.625 \end{gathered}$$

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" $$\begin{gathered} \frac{3-{\displaystyle \frac{1}{2}}}{mc}=1.6 \\ mc=4 \end{gathered}$$

By using the coating's index of refraction in equation is $$\begin{gathered} n=\frac{400\times 4}{2\times 560} \\ n=1.429 \end{gathered}$$

Centering is$\boxed{n=1.429}$