Subject: Applied Physics 2

Topic: Interference And Diffraction

Difficulty: Medium

A light beam is incident at an angle of $40^0$ with normal on a plane parallel film of thickness $4 \times 10^{-5}$ cm & RI 1.5. Show that $\lambda = 7.228 \times 10^{-5} $cm will be strengthened in the reflected light.

t=4 × 10${-5}$ cm , i=40˚, μ = 1.5

μ= $\frac{sin\ i}{sin\ r}$

sinr = $\frac{sin \ i}{μ}$

sin r = 0.42

Therefore r= 25.37

cos r = 0.9035

For constructive interference in reflected light 2μ t cos r = (2n±1) $\frac{λ}{2}$

2μ t cos r = m $\frac{λ}{2}$ , where m= (2n±1)

Substitute all the values in the equation if m turns out to be odd number then the wavelength will be strengthened.

$ 2 \times 1.5 \times 0.9035 = m \times 7.228 \times \frac{10^{-5}}{2} $

Therefore m= 5 hence the wavelength $λ = 7.228 × 10^{-5}$ cm will be strengthened in the reflected light.