Question

To produce a minimum reflection of wavelengths near the middle of visible spectrum ($550nm$), how thick should a coating of $Mg{F}_2$ ($\mu =1.38$) be vacuum-coated on a glass surface?

• A. ${10}^{-7}$
• B. ${10}^{-10}$
• C. ${10}^{-9}m$
• D. ${10}^{-8}m$

Answer 1

The correct option is A ${10}^{-7}$

Concept:-Approach to this kind of problems are like first find ${\lambda }_{film}$which equal to $\frac{\lambda }{\mu }$; here $\lambda$ is the wavelength of light to be minimized reflection & $\mu$ is refractive index of coating material.

${\lambda }_{film}=\frac{\lambda }{\mu }$.

Then the thickness is equal to $\frac{{\lambda }_{film}}{\mu }$

i.e t=$\frac{{\lambda }_{film}}{\mu }$

Solution:- Given $\mu =1.38$ & $\lambda =550nm$

Hence ${\lambda }_{film}=\frac{\lambda }{\mu }=\frac{550}{1.38}$

${\lambda }_{film}=398.55nm$

Now,thickness of coating ,t is.

$t=\frac{{\lambda }_{film}}{\mu }=\frac{398.55nm}{\mu }$

$t\approx 100nm\approx {10}^{-7}m$