Lloyd’s Mirror Experiment, Interference In a Thin Film (JEE LEVEL)

Let the thickness of the film be t and refractive index be \displaystyle \mu (> 1). At A it is partly reflected along \displaystyle A{{R}_{1}} giving the ray (1) and partly refracted along AB at an angle r. At B it is again partly reflected along BC and partly refracted along \displaystyle B{{T}_{1}}. Similar reflections and refractions occur at C. Since, the rays \displaystyle A{{R}_{1}} and \displaystyle C{{R}_{2}}, i.e. ray (1) and ray (2) have been derived from the same incident ray, they are coherent and in a position to interfere. If the set of parallel reflected rays is now collected by a lens, and focused at P, each ray has traveled a different distance, and the phase relationship between them may be such as to produce destructive or constructive interference at P. It is such interference that produces the colors of this film when seen by naked eyes.

         Let CN and BM be perpendiculars to \displaystyle A{{R}_{1}} and AC. As the paths of the rays \displaystyle A{{R}_{1}} and \displaystyle C{{R}_{2}} beyond CN are equal, the path difference between ray (1) and (2) is given by

(path ABC in film-path AN in air)

therefore  path difference = \displaystyle \mu  (AB + BC) – AN    … (1)

Here     \displaystyle AB=BC=\frac{{BM}}{{\cos r}}=\frac{t}{{\cos r}},

and       AN = AC sin i

Now      AC = AM + MC = BM tan r + BM tan r

                   = 2t tan r

Also      AN = 2t tan r sin i

                    \displaystyle =2t\frac{{\sin r}}{{\cos r}}(\sin i)

                    \displaystyle =2t\frac{{\sin r}}{{\cos r}}(\mu \sin r)        

                                  \displaystyle \left[ {\because \ \ \frac{{\sin i}}{{\sin r}}=\mu } \right]

                    \displaystyle =2\mu t\frac{{{{{\sin }}^{2}}r}}{{\cos r}}

Substituting these values of AB, BC and AN in Eq. (1) we get path difference

    \displaystyle =\mu \left( {\frac{t}{{\cos r}}+\frac{t}{{\cos r}}} \right)-2\mu t\frac{{{{{\sin }}^{2}}r}}{{\cos r}}

\displaystyle =\frac{{2\mu t}}{{\cos r}}(1-{{\sin }^{2}}r)

hence path difference  = 2 \displaystyle \mu  t cos r

However, we must take account of the fact that ray (1) undergoes a phase change of \pi at reflection while ray (2) does not, since it is internally reflected . The phase change of \pi  is equivalent to a path difference of \displaystyle \frac{\lambda }{2}. Hence, the effective path difference between ray (1) and rays (2) is

\displaystyle 2\mu t\cos r-\frac{\lambda }{2}

Now for constructive interference this path difference must be in integral multiple of \displaystyle \lambda . Hence

\displaystyle \begin{array}{l}\text{       }2\mu t\cos r-\frac{\lambda }{2}=n\lambda \\\Rightarrow \text{  }2\mu t\cos r=n\lambda +\frac{\lambda }{2}\end{array}

or     \displaystyle 2\mu t\cos r=(2n+1)\frac{\lambda }{2}

So this becomes a condition for constructive interference as far as reflected rays ( \displaystyle {{R}_{1}}) and ( \displaystyle {{R}_{2}}) are concerned.

Whereas for transmitted rays \displaystyle {{T}_{1}} and \displaystyle {{T}_{2}} there is only one path difference between them which is   \displaystyle 2\mu t\cos r and there is no extra path difference due to reflection.

Hence for constructive interference this path difference must be in integral multiple of wavelength \displaystyle \lambda . Hence

\displaystyle 2\mu t\cos r=n\lambda

So this becomes a condition for constructive interference as far as transmitted rays ( \displaystyle {{T}_{1}}) and ( \displaystyle {{T}_{2}}) are concerned.

         Thus, when a thin film is illuminated by monochromatic light, and seen in reflected light, it appears bright or dark according to the path difference which is given as 2 \displaystyle \mu tcos(r) .

\displaystyle 2\mu t\cos r=(2n+1)\frac{\lambda }{2} (condition of maxima)

\displaystyle 2\mu t\cos r=n\lambda   (condition of minima)

Conditions of Maxima and Minima for transmitted light

The rays emerging from the lower side of the film can also be brought together with a lens and made to interfere.

The path difference between the transmitted rays \displaystyle B{{T}_{1}} and \displaystyle D{{T}_{2}} is given by

 (BC + CD) – BL = 2 \displaystyle \mu t cos r

In this case, film there are no phase change due to reflection at B or C, because in either case the light is traveling from denser to rarer medium. Hence, the effective path difference between \displaystyle B{{T}_{1}} and \displaystyle B{{T}_{2}} is also 2 \displaystyle \mu t cos r.

The two rays \displaystyle B{{T}_{1}} and \displaystyle D{{T}_{2}}  reinforce each other, if 2 \displaystyle \mu  t cos r = n \displaystyle \lambda  (condition of maxima) where n = 1, 2, 3.

In this case, the film will appear bright in the transmitted light.

The two rays will destroy each other if

\displaystyle 2\mu t\cos r=(2n+1)\frac{\lambda }{2}  (condition of minima)

where n = 0, 1, 2,…. and the film appears dark in transmitted light.

         Hence we find that the conditions for the maxima and minima, in the reflected light are just the reverse of those in transmitted light. Therefore, only those colors will be visible in transmitted light, which were missed in reflected light. Hence, the film which appears bright in reflected light will appear dark in transmitted light and vice-versa. In other words, the appearances of colors in the two cases is complimentary to each other.

Reason of Colors in Reflection from a Thin Films

The sunlight consists of a continuous range of wavelengths (colors). At a particular point of the film, and for a particular position of the eye (i.e., for a particular t and a particular r), the rays of only certain wavelengths will have a path difference satisfying the condition of maxima. Hence, only these wavelengths (colors) will be present with the maximum intensity. While some others, which satisfy the condition of the minimum will be missing. Hence, the point of the film being viewed will appear colored.

Illustration

A thin film of \displaystyle 4\times {{10}^{{-5}}} cm thickness is illuminated by white light normal to its surface (r = \displaystyle 0{}^\circ ). Its refractive index is 1.5. Of what color will the thin film appear in reflected light? (Visible range 3500 – 7000 \displaystyle {{A}^{o}} )

Solution

The condition for constructive interference of light reflected, from a film is

\displaystyle 2\mu t\cos r=(2n+1)\frac{\lambda }{2},\ \text{where}\ n=0,1,2,….

Here \displaystyle \mu =1.5,

\displaystyle t=4\times {{10}^{{-5}}}\text{cm}\ \text{and}\ r=0{}^\circ (since light falls normally) so that cos r = 1.

therefore   2 × 1.5 × 4 × \displaystyle {{10}^{{5}}}= (2n + 1) \displaystyle \frac{\lambda }{2}

or     \displaystyle \lambda =\frac{{4\times 1.5\times 4\times {{{10}}^{{-5}}}}}{{2n+1}} cm

or    \displaystyle \lambda =\frac{{24\times {{{10}}^{{-5}}}}}{{2n+1}} cm

or  \displaystyle \lambda =\frac{{24000}}{{2n+1}}{{A}^{o}}

Taking n = 0, 1, 2, 3, ….. we get different wavelengths. 

For n=0 , \displaystyle \lambda =24000{{A}^{o}}

For n=1 , \displaystyle \lambda =8000{{A}^{o}}

For n=2 , \displaystyle \lambda =4800{{A}^{o}}

For n=3 , \displaystyle \lambda =3428{{A}^{o}}

These are the wavelengths reflected most strongly. Of these, the wavelength lying in the visible region is 4800 \displaystyle {\AA} (blue).

Interference by a Wedge-Shaped Film

Considering the interference pattern produced by a film of varying thickness, i.e., a film which is not plane-parallel. Observe that the interfering rays do not enter the eye parallel to each other but appear to diverge from a point near the film.