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?


Answer:

6.71×10⁻⁷ m

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


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