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, ${\lambda }_1$ = 700 nm and ${\lambda }_2$ = 420 nm will be weekly reflected for both wavelengths is ${10}^{-7}$m. Find x ?

Answer 1

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

$⟹d=\frac{(2n-1)\lambda }{4\mu }$

For light of wavelength ${\lambda }_1=$ 700 nm

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

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

$d_2=\frac{420}{4\mu },\frac{1260}{4\mu },\frac{2100}{4\mu }$

$⟹d_{min}=\frac{2100}{4\mu }=\frac{2100}{4\times \frac{4}{7}}$

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