Chapter 24, Problem 42P

To determine

The minimum thickness of coating.

Answer to Problem 42P

  $228nm$

Explanation of Solution

Given info:

Wavelength, $\lambda =570nm$

The refractive index, $n=1.25$

Formula used:

Dark bands will occur when $2t=m\lambda ,m=0,1,2,\mathrm{...}$

The light rays are perpendicular to the plates, the extra path length equals $2t$ , where $t$ is the thickness of the air gap at any point.

Where,

  $m$ is an integer.

  $t$ is the thickness of the film.

  $\lambda$ is wavelength.

Calculation:

By using the formula, $t=\frac{1}{2}m\lambda$

For constructive interference $m=1$

Due to a phase shift of $\frac{\lambda }{2}$ the equation becomes $t=\frac{\lambda }{2n}=\frac{570}{2\left(1.25\right)}=228nm$

  $\lambda =570nm$ is in the middle of the visible spectrum.

The transmitted light will be stronger in the wavelengths at the ends of the spectrum.

So, the lens would emphasize the red and violet wavelengths.

Conclusion: The required thickness is $228nm$ .