Question

To reduce the light reflected by the glass surface of a camera lens, the surface is coated with a thin layer of another material which has an index of refraction $(\mu =7/4)$ smaller than that of glass. The least thickness of the layer, to ensure that light falling perpendicularly on the surface and having wavelengths, ${\lambda }_1=700nm$ and ${\lambda }_2=420nm$ will be weekly reflected for both wavelengths is $x\times {10}^{-8}m$. Find x ?

Answer 1

path difference, $2\mu d=(2n-1)\frac{\lambda }{2}(n=1,2,3,.....)$

$\Rightarrow d=\frac{(2n-1)\lambda }{4\mu }$

For light of wavelength

${\lambda }_1=700nm$

$d_1=\frac{700}{4\mu },\frac{2100}{4\mu },\frac{3500}{4\mu }$

for light of wavelength ${\lambda }_2=420nm$

$d_2=\frac{420}{4\mu },\frac{1260}{4\mu },\frac{2100}{4\mu }\Rightarrow d_{min}=\frac{2100}{4\mu }=\frac{2100}{4\times \frac{7}{4}}$

$d_{min}=300nm=3\times {10}^{-7}m=30\times {10}^{-8}m$