Coating in Lenses!

Classical Mechanics Level 5

To ensure almost $100$ % 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 ( $d$ in Angstrom) be so that at the center of the visible spectrum $(5500 Angstrom)$ , there is maximum transmission?

Note : Round off the value of $d$ to the nearest 1000. (ex. if the value comes out to be 5997, then the answer should be 6000)

Source - NCERT Exemplar Problems

The answer is 1000.

2 solutions

Nishant Rai
Jun 3, 2015

Tanishq Varshney
Jun 3, 2015

hope u r reading this

Path difference between ray $1$ and $2$

$\Delta x=2\mu _{film} d \cos r+\frac{\lambda}{2}$

path difference between ray $3$ and $4$

$\Delta x=2\mu_{film} d \cos r$

$r$ is angle of refraction.

For maximum transmission $cos r=1$ so $r=0^{o}$ and of course $\Delta x=n \lambda$

Shouldn't the case of $3$ and $4$ be considered during transmission.

so by this we get

$2\mu _{film} d =\lambda$ for $n=1$

$d= \frac{5500}{2\times 1.38}$

$d= 1992.7~angstrom$ and according to question rounding off gives 2000

but the answer you posted, is get by using path difference between $1$ and $2$

Why is it so, plz correct me if i am wrong and do reply

@Nishant Rai do reply

Tanishq Varshney - 6 years ago

@Tanishq Varshney Check the solution i have posted. This is what NCERT has provided.

Nishant Rai - 6 years ago

ok but plz do consider by statement , see the diagram above i posted , during transmission ray 3 and 4 will come out, correct??

Tanishq Varshney - 6 years ago

@Tanishq Varshney – yes, you are correct.

@Rohit Gupta Please Help!

Nishant Rai - 6 years ago