Interference by Reflected Rays from a Wedge-Shaped Thin Film

The two surfaces of the thin film are slightly inclined to each other at an angle \( \theta \). Therefore, the thickness of the film gradually increases from one end to the other.

The ray \(SA\) is the incident ray falling on the upper surface of the thin film at \(A\). At \(A\), the incident light is partially reflected and partially transmitted.

A part of the incident light is reflected directly from the upper surface at \(A\) and travels along \(AR_1\). This is the first reflected ray.

The remaining part of the incident ray enters the thin film at \(A\). It travels inside the film and reaches the lower surface at \(B\).

At \(B\), a part of the light is reflected back towards the upper surface. It reaches the upper surface at \(C\) and emerges along \(CR_2\).

Therefore, the two rays which interfere are: \(R_1\) and \(R_2\).

The rays \(R_1\) and \(R_2\) emerge in nearly the same direction. Therefore, they are coherent rays and can interfere with each other.

The interference depends upon their optical path difference. The ray \(R_2\) travels an additional distance inside the thin film \(AB\) + \(BC\)

Thus

\[\Delta_1 = \mu (AB + BC) - AD .....(1) \]

where:

• \(\mu\) = refractive index of the thin film
• \(CD\) = Normal on \(AR_1\) from \(C\)

Consider the two triangles \( \triangle ACD \) and \( \triangle ACG \).

For \( \triangle ACD \)

\( i + \angle DAC = \angle DAC + \angle ACD = 90^\circ \)

Therefore,

\( \angle ACD = i \)

For \( \triangle ACG \)

\( r + \angle CAG = \angle CAG + \angle ACG = 90^\circ \)

Therefore,

\( \angle ACG = r \)

Using Snell's Law

The refractive index of the film is given by:

\[ \mu = \frac{\sin i}{\sin r} \]

From the geometry of the triangles (\( \triangle ACD \) and \( \triangle ACG \)):

\[ \sin i = \frac{AD}{AC} \]

\[ \sin r = \frac{AG}{AC} \]

Hence,

\[ \mu = \frac{\sin i}{\sin r} = \frac{AD/AC}{AG/AC} = \frac{AD}{AG} \]

Therefore,

\[ \boxed{AD = \mu \, AG}.......(2) \]

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To determine the length of the path of the light ray \(BC\), draw a perpendicular \(CE\) from point \(C\) to the lower surface of the film.

The perpendicular \(CE\) and the light ray \(AB\), when produced forward, meet at point \(F\).

The exterior angle of a triangle is equal to the sum of its two interior opposite angles.

\[ \angle ABE = \angle MBA + 90^\circ \]

Since,

\[ \angle ABE = \theta + 90^\circ + r \]

Therefore,

\[ \angle MBA = \theta + r \]

Since the angles formed by the intersecting lines are equal:

\[ \theta + r = \angle MBA = \angle M'BF = \angle BFC \]

and

\[ \angle MBC = \angle BCF \]

Therefore, \( \triangle BCF \) is an isosceles triangle.

Hence, the two opposite sides are equal:

\[ {BC = BF} \]

Also, since \(CE\) and \(EF\) are equal:

\[ {CE = EF = t}.......(3) \]

On substituting equations (2) and (3) into equation (1), we get:

\[ \Delta_1 = \mu(AB + BF - AG) \]

Since \(AB + BF - AG = FG\), therefore:

\[ \Delta_1 = \mu FG \]

From the geometry of the figure \( \triangle CGF \):

\[ FG = CF \cos(\theta + r) \]

Therefore:

\[ \Delta_1 = \mu CF \cos(\theta + r) \]

Since:

\[ CE = EF = t \]

and \( \triangle CEF \) is an isosceles triangle, we have:

\[ CF = 2t \]

Hence, the optical path difference becomes:

\[ \boxed{ \Delta_1 = 2\mu t\cos(\theta+r) } \]

Stokes's Law

When light is reflected from a rarer medium to a denser medium, it undergoes a phase change of \( \pi \) radians. This is equivalent to an additional path difference of \({\frac{\lambda}{2}} \)

Hence, the total path difference between the two reflected rays is:

\[ \boxed{ \Delta = 2\mu t\cos(\theta+r)+\frac{\lambda}{2}} \]

Case (i) Constructive interference:

\[ \Delta=n\lambda \]

Therefore:

\[ 2\mu t\cos(\theta+r)+\frac{\lambda}{2}=n\lambda \]

or:

\[ \boxed{ 2\mu t\cos(\theta+r)= \left(2n-1\right)\frac{\lambda}{2} } \] where, n=1,2,3,....

Case (ii) Destructive interference

For :

\[ \Delta = \left(2n+1\right)\frac{\lambda}{2} \]

Therefore:

\[ 2\mu t\cos(\theta+r)+\frac{\lambda}{2} = \left(2n+1\right)\frac{\lambda}{2} \]

Hence:

\[ \boxed{2\mu t\cos(\theta+r)=n\lambda} \] where, n=0,1,2,3,....