Question

a material having and index of refraction of 1.30 is used as an anti-reflective coating on a piece of glass (n=1.50). what should the minimum thickness of the thin film be to minimize reflection of 500-nm light

Answer 1

Concepts and reason

This question is based upon the concept of the destructive interference.

Firstly, write the equation for destructive interference. Rearrange the equation for thickness and calculate the value of minimum thickness of the thin film. Destructive interference occurs

Fundamentals

The expression for destructive interference is,

$2nt = \left( {m + \frac{1}{2}} \right)\lambda$

Here, $\lambda$ is the wavelength, $m$ is the order number, $t$ is the thickness of film and $n$ is the index of the refraction.

The expression for destructive interference is,

$2nt = \left( {m + \frac{1}{2}} \right)\lambda$

Rearrange the equation for thickness of non-reflecting film .

$\begin{array}{c}\\2nt = \left( {m + \frac{1}{2}} \right)\lambda \\\\t = \frac{{\left( {m + \frac{1}{2}} \right)\lambda }}{{2n}}\\\end{array}$

Minimum value of $m$ is zero.

Substitute $0$ for $m$ in the above equation.

$\begin{array}{c}\\t = \frac{{\left( {0 + \frac{1}{2}} \right)\lambda }}{{2n}}\\\\t = \frac{\lambda }{{4n}}\\\end{array}$

The expression for thickness of non-reflecting film is,

$t = \frac{\lambda }{{4n}}$

Substitute $500{\rm{ nm}}$ for $\lambda$ and $1.3$ for $n$ in above equation of height.

$\begin{array}{c}\\t = \frac{{500{\rm{ nm}}}}{{4\left( {1.3} \right)}}\\\\ = 96{\rm{ nm}}\\\end{array}$

Ans:

The value of thickness of non-reflecting film is $96{\rm{ nm}}$ .