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

detailed solution

Correct option is A

Consider light to be incident at near normal incidence. We wish to cause destructive interference between rays r1 and r2 so that maximum energy passes into the glass. A phase change of λ/2 occurs in each ray because at both the upper and lower surfaces of the MgF2 film the light is reflected by a medium of greater index of refraction. When striking a medium of lower index of refraction, the light is reflected with no phase change. Since in this problem both rays I and 2 experience the same phase shift, no net change of phase is introduced by these two reflections. Hence, the only way a different optical path lengths, the optical path length is the product of the geometric path difference a ray travels through different media and the refractive index of the medium in which it is travelling. For destructive interference. The two rays must be out of phase by an odd number of half wavelengths. Hence the optical path difference needed for destructive interference is 2μd=(2n+1)λ2,n=0,1,2....... Note that 2μd is the total optical path length that the raysTraverse when n=0∴d=λ/22μ=λ4μ=350×10−94×1.38=100nm=1×10−7m