To ensure almost 100 per cent transmittivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of air and glass (which makes the optical element of the lens). A typically used dielectric film is `MgF_(2)(n=1.38)`. What should the thickness of the film be so that at the center of the visible speetrum `(5500Å)` there is maximum transmission.

Explanation

As shown in the figure, suppose a ray of light `vecPA` is incident on the surface `M_(1)N_(1)` of dielectric layer at angle of incidence i at time t = 0. As a result, we obtain `vecAQ=vecr_(1)` as reflected ray and `vecAD` as refracted ray. Further ray `vecAD` gets reflected from surface `M_(2)N_(2)`, to give reflected ray as `vecDC` which gets transmitted from point C to give ray `vecCR=vecr_(2)`. Here incident ray `vecPA` undergoes successive reflections and transmissions and so its amplitude goes on decreasing. Hence, intensity of light at point A is majority due to light rays `r_(1)` and `r_(2)` is,
Now, optical path difference between light rays `r_(1)` and `r_(2)` is,
`r_(2)-r_(1)=n(AD)+n(DC)-AB" "......(1)`
In right angled `DeltaAED,`
`cosr=(d)/(AD)impliesAD=(d)/(cosr)`
In right angled `DeltaDEC`,
`cosr=(d)/(DC)impliesDC=(d)/(cosr)`
Now in right angled `DeltaABC`,
`sini=(AB)/(AC)=AB=(AC)sini`
In right angled `DeltaAED,tanr=(((AC)/(2)))/(d)=(AC)/(2d)`
`:.AC=2dtanr`
`:.AB=(2dtanr)sini`
Placing above values in equation (1), optical path difference is,
`r_(2)-r_(1)=n((d)/(cosr))+n((d)/(cosr))-(2dtanr)sini`
`=(2nd)/(cosr)-2d((sinr)/(cosr))(nsinr)("":.n=(sini)/(sinr))`
`:.r_(2)-r_(1)=(2nd)/(cosr)(1-sin^(2)r)`
`=(2nd)/(cosr)(cos^(2)r)`
`r_(2)-r_(1)=2ndcosr" ".......(2)`
Here both the rays `(r_(1) and r_(2))` get reflected from , the surfaces of denser transparent medium and so there is no additional path difference between them.
Now, in order to have maximum transmissio through the layer of `MgF_(2)`, there must not be ar reflection from its upper surface `M_(1)N_(1)`. In othe words, destructive interference must take plac between `r_(1)` and `r_(2)`. For this, above path difference should be equal to `(lamda)/(2)` (for minimum thickness of layer). Hence,
`2ndcosr=(lamda)/(2)`
Here light rays are made incident normall (perpendicularly) on the lens and so
`i=r=0^(@)impliescosr=1` and so,
`:.2nd=(lamda)/(2)`
`:.d=(lamda)/(4n)`
`:.d=(5500)/(4xx1.38)`
`=996.4Å`
`:.d~~1000Å`