41.5k views
0 votes
A thin layer of oil of refractive index 1.22 is spread on the surface of water (n = 1.33), If the thickness of the oil is 275 nm, then what is the wavelength of light in air that will be predomínantly reflected from the top surface of the oil?

User Apgsn
by
8.4k points

1 Answer

3 votes

Answer:

6.71×10⁻⁷ m

Step-by-step explanation:

Using thin film constructive interference formula as:

2×n×t = m×λ

Where,

n is the refractive index of the refracted surface

t is the thickness of the surface

λ is the wavelength

If m =1

Then,

2×n×t = λ

Given that refractive index pf the oil is 1.22

Thickness of the oil = 275 nm

Also, 1 nm = 10⁻⁹ m

Thickness = 275×10⁻⁹ m

So,

Wavelength is :

λ= 2×n×t = 2× 1.22 × 275×10⁻⁹ m = 6.71×10⁻⁷ m

User Denis Malinovsky
by
8.2k points