A lens (μ=1.5) is coated with a thin film of refractive index 1.2 in order to reduce the reflection from its surface at λ=4800Ao. The minimum thickness of the film which will minimize the intensity of the reflected light is n(10−7) meter. Then n is

Path difference between two reflected ray in air,Δx=2μ1t+(λ2−λ2)=2μ1t For the minimum intensity, Δx=(2n−1)λ2 ⇒2μ1t=(2n−1)λ2t=(2n−1)λ4μ1=λ4μ1(For minimum thickness, n=1)t=λ4μ1=4800×10−104×1.2=10−7m

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A lens $(μ = 1.5)$ is coated with a thin film of refractive index 1.2 in order to reduce the reflection from its surface at $λ = 4800 \overset{o}{A}$ . The minimum thickness of the film which will minimize the intensity of the reflected light is $n (10^{- 7})$ meter. Then n is

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Detailed Solution

Path difference between two reflected ray in air,

$\begin{array}{l} Δ x = 2 μ_{1} t + (\frac{λ}{2} - \frac{λ}{2}) = 2 μ_{1} t \\ F o r t h e \min i m u m int e n s i t y, Δ x = (2 n - 1) \frac{λ}{2} \\ \Rightarrow 2 μ_{1} t = (2 n - 1) \frac{λ}{2} \end{array}$

$\begin{array}{l} t = \frac{(2 n - 1) λ}{4 μ_{1}} = \frac{λ}{4 μ_{1}} (F o r \min i m u m t h i c k n e s s, n = 1) \\ t = \frac{λ}{4 μ_{1}} = \frac{4800 \times 10^{- 10}}{4 \times 1.2} = 10^{- 7} m \end{array}$