Question

A diver is swimming underneath anoil slickwith a thickness of 200nm and an index of refraction of 1.50. A white light shines straight down towards the diver from above the oil slick. The index of refraction of water is 1.33.What is the longest wavelengthof the light in water, 휆푤푎푡푒푟,that is transmitted most easily to the diver?

Answer 1

Answer:

$\lambda_{water} = 451.13\ nm$

Explanation:

given,

Thickness of oil = 200 nm

refractive index of oil (n₁)= 1.50

refractive index of water (n₂) = 1.33

$n_{air} \lambda_{air} = n_{oil} \lambda_{oil}$

Wavelength of light in air medium

2t = k\lambda_{oil}

k = 1

$2t = \dfrac{n_{air} \lambda_{air}}{n_{oil}}$

$2t = \dfrac{n_{air} \lambda_{air}}{n_{oil}}$

$\lambda_{air} = \dfrac{2t\ n_{oil}}{n_{air}}$

$\lambda_{air} = \dfrac{2\times 200 \times 1.5}{1}$

$\lambda_{air} = 600\ nm$

now,

wavelength of light in water

$\lambda_{water} = \dfrac{n_{air} \lambda_{air}}{n_{water}}$

$\lambda_{water} = \dfrac{1  \times 600 nm}{1.33}$

$\lambda_{water} = 451.13\ nm$