A thin film of a specific material can be used to decrease the intensity of reflected light. There is destructive interference of waves reflected from upper and lower surfaces of the film. These films

Question

A thin film of a specific material can be used to decrease the intensity of reflected light. There is destructive interference of waves reflected from upper and lower surfaces of the film. These films are called anti-reflection coating. Magnesium fluoride $(MgF_{2})$ is used as anti-reflecting coating on plane glass surface for light having wavelength 500 nm and normal incidence. The minimum thickness of film required is 100 nm. Find refractive index of magnesium fluoride. It is known that refractive index of glass is more than refractive index of magnesium fluoride.

Answer:

1.25

Solution:

Since $n_{g l a ss} > n_{f i l m} > n_{ai r}$ , both reflected waves undergo equal phase reversal, so the path difference alone must give destructive interference.

Minimum film: $2 μ t = \frac{λ}{2}$

$μ = \frac{λ}{4 t} = \frac{500}{4 \times 100} = 1.25$

TestHub · Accepted Answer

Since n g l a ss ​ > n f i l m ​ > n ai r ​ , both reflected waves undergo equal phase reversal, so the path difference alone must give destructive interference. Minimum film: 2 μ t = 2 λ ​ μ = 4 t λ ​ = 4 × 100 500 ​ = 1.25

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