Interference by transmitted rays

Consider a plane-parallel transparent thin film of refractive index μ and thickness t. A ray of monochromatic light is incident on the upper surface of the film at an angle of incidence i. A part of the incident ray is reflected from the upper surface, while the remaining part is refracted into the film.

let draw two perpendicular $C_2B$ and $P_2M$. then effective pathdifference

$$\Delta_1 = \mu(C_1P_2 - P_2C_2)- C_1B \qquad ...(1)$$ $\because \Delta C_1MP_2 \approx \Delta C_2MP_2$ $$\therefore C_1P_2 = P_2C_2 = \frac{t}{\cos r} \qquad ...(2)$$ $$ and \qquad C_1M = MC_2 = t\tan r \qquad ...(3)$$

In $\Delta C_1AC_2$

$$\sin i = \frac{C_1A}{C_1C_2}= \frac{C_1A}{C_1N + NC_2}$$ Hence, $$C_1A = 2t \sin i \tan r \qquad..(4) \qquad \because eq.(3)$$ Thus by eq. (1), (3) and (4)

$$\Delta_1 = \mu(\frac {2t}{\cos r})- 2t \sin i \tan r $$ $$\Delta_1 = 2\mu t (\frac {1}{\cos r}- \frac {\sin i \tan r}{\mu} ) $$ $$\Delta_1 = 2\mu t (\frac {1}{\cos r}- \frac {\sin i \sin r}{\mu \cos r} ) $$ $$\Delta_1 = 2\mu t (\frac {1}{\cos r}- \frac {\sin i {\sin^2 r}}{\sin i \cos r} ) $$ $$\Delta_1 = 2\mu t (\frac {1 - \sin^2 r}{\cos r} $$ $$\Delta_1 = 2\mu t \cos r ) $$

Case (1) Constructive Interference

For constructive interference, the total optical path difference must be an integral multiple of the wavelength:

$$\Delta = n\lambda$$

Therefore, the condition for dark or destructive interference in reflected light is:

$$\boxed{2\mu t\cos r = n\lambda }$$ $$where, \qquad n = 0,1,2,3,... $$

Case (2) Destructive Interference

For destructive interference, the total path difference must be an odd multiple of half a wavelength: $$\Delta = \left(2n+1\right)\frac{\lambda}{2}$$

Therefore, the condition for bright or constructive interference in reflected light is: $$\boxed{ 2\mu t\cos r = \frac{(2n+1)\lambda}{2}} $$ $$where, \qquad n = 0,1,2,3,... $$