Question

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 $Mg{F}_2$ (n = 1.38). What should the thickness of the film be so that at the center of the visible spectrum ($5500\stackrel{o}{A}$) there is maximum transmission.

• A. $5000\stackrel{o}{A}$
• B. $2000\stackrel{o}{A}$
• C. $1000\stackrel{o}{A}$
• D. $3000\stackrel{o}{A}$

Answer 1

The correct option is C $1000\stackrel{o}{A}$

Consider a ray incident at angle i. A part of this ray is reflected from the air-film interface and a part refracted inside. This is partly reflected at the film-glass interface and a part transmitted. A part of the reflected ray is reflected at the film-air interface and a part transmitted as $r_2$ parallel to $r_1$. Of course, successive reflections and transmissions will keep on decreasing the amplitude of the wave. Hence rays $r_1$ and $r_2$ shall dominate the behavior. If incident light is to be

transmitted through the lens, $r_1$ and $r_2$ should interfere destructively. Both the reflections at A and D are from lower to higher refractive index and hence there is no phase change on reflection. The optical path difference between $r_2$ and $r_1$ is n(AD +CD) - AB.

If d is the thickness of the film , then

$AD=CD=\frac{d}{cosr}$

$AB=ACsini$

$\frac{AC}{2}=dtanr$

$\therefore AC=2dtanr$

Hence, $AB=2dtanr\times sini$

Thus the optical path difference is

$2n\frac{d}{cosr}2dtanr\times sini$

= $2\cdot \frac{sini}{sinr}\cdot \frac{d}{cosr}\cdot 2d\cdot \frac{sinr}{cosr}\cdot sini$

=$2d\cdot sini\left[\frac{1-si{n}^{2}r}{sinrcosr}\right]=2nd\cdot cosr$

For these waves to interfere destructively this must be $\lambda$ \2.

$2ndcosr=\frac{\lambda }{2}$ ot $ndcosr=\frac{\lambda }{4}$

For a camera lens, the sources are in the vertical plane and hence

$i\simeq r\simeq 0$

$\therefore nd\simeq \frac{\lambda }{4}$

$\Rightarrow$ $d=\frac{5500\stackrel{0}{A}}{1.38\times 4}\simeq 1000\stackrel{0}{A}$