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

The minimum thickness of the oil is 77.55 nm

Step-by-step explanation:

Given:

Refractive index of oil
`n_(o) = 1.47`

Refractive index of water
`n_(w) = 1.35`

Wavelength of light
`\lambda= 456 * 10^(-9)` m

From the equation of thin film interference,

The minimum thickness is given by,

`2n_(o) t = (n+(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+(1)/(2) )\lambda`

`t = (\lambda )/(4 n_(o) )`

`t = (456 * 10^(-9) )/(4 * 1.47)`

`t = 77.55 * 10^(-9)` m

`t = 77.55` nm

Therefore, the minimum thickness of the oil is 77.55 nm