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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. |
Answer» Solution : 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` |
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