Question

A thin film of cooking oil ( n = 1.47 ) is spread on a puddle of water ( n = 1.35 ) . What is the minimum thickness D min of the oil that will strongly reflect blue light having a wavelength in air of 456 nm, at normal incidence?

Answer 1

Answer:

The minimum thickness of the oil is 77.55 nm

Explanation:

Given:

Refractive index of oil $n_{o} = 1.47$

Refractive index of water $n_{w} = 1.35$

Wavelength of light $\lambda= 456 \times 10^{-9}$ m

From the equation of thin film interference,

The minimum thickness is given by,

    $2n_{o} t = (n+\frac{1}{2}) \lambda$

Where $n = 0,1,2,3.........$,$t =$ thickness

Here we have to find minimum thickness so we use $n = 0$

     $2n_{o} t =( 0+\frac{1}{2} )\lambda$

   $t = \frac{\lambda }{4 n_{o} }$

   $t = \frac{456 \times 10^{-9} }{4 \times 1.47}$

   $t = 77.55 \times 10^{-9}$ m

   $t = 77.55$ nm

Therefore, the minimum thickness of the oil is 77.55 nm