Complete destructive interference of reflected light is observed for $5000A^o$ and $7000A^o$ and for no other in between. Find thickness of oil layer $n_{oil} = 1.3$ , $n_{glass} = 1.5$ .

Mumbai university > FE > SEM 2 > Applied Physics II

Marks: 7M

Year: May 2013

Given:

$n_o = 1.3$

$n_g = 1.5$

$λ_1 = 5 × 10^{-7}m = λ_{n+1}$ ...(No wavelength between estructive interference)

them show

$λ_2 = 7 × 10^{-7}m = λ_n $

To find:

t

Solution:

The set u is similar to anti-reflective coating.

For destructive interference

$2 μt cos r = (2n - 1)\frac{λ_n}{2} = [2(n + 1) -1] \frac{λ_{n+1}}{2} ....(i)$

$(2n - 1) \frac{5 × 10^{-7}}{2} = (2n + 1) \frac{7 × 10^{-7}}{2}$

$n = 3.....(ii)$

For, normal incidence, cos r =1,

from (i) and (ii)

$2 μt = (2 × 3 - 1) \frac{λ_n}{2} = \frac{5 × 5 × 10^{-7}}{2}$

$t = \frac{12.5 × 10^{-7}}{2 × μ} = \frac{12.5 × 10^{-7}}{2 × 1.3}.......(μ = n_{oil})$

$t = 6.73 × 10^{-7}m \ \ or \ \ t = 0.673 μm $