Winter 2016 – Q.3

3. (a) What is the fundamental principle of a Hologram? How is it produced and how the image constructed by it ? [8M]

Holography is the technique of production of a three dimensional image of an object. In general photography only intensity is recorded to produce the two dimensional image of an object. In holography both the intensity and phase of the wave are recorded and when viewed the photograph shows a three dimensional image of the object.

The process of hologram recording is shown in the figure. A weak but broad beam of laser light is split into two beams namely reference beam and object beam. The reference beam is allowed to reach the photographic plate directly while the object beam illuminates the object. Part of the light scattered by the  object travels towards the photographic plate and interferes with the reference beam and produces an interference pattern on the photographic plate. The photographic plate that carries the interference pattern is called as hologram. The hologram then can be developed and fixed.

A laser beam identical to the reference beam is used for the reconstruction of an object. The
reconstruction beam illuminates the hologram at the same angle as the reference beam. The hologram acts as the diffraction grating and the secondary waves from the hologram interfere constructively in certain directions and interfere destructively in other directions. They form a real image in front of the hologram and virtual image behind the hologram at original site of the object. An observer sees light waves diverging from the virtual image. An image of the object appears where the object once stood and that image is identical to what our eyes would have perceived in all its details. If the observer tilts his head other objects behind the first one or new details objects which were not noticed earlier would be observed.

If sunlight is incident on a soap film, the optical path difference will vary from one color to other due to change of wavelength with color. Therefore the film will appear to be colored and the color seen will be of that rays for which constructive interference occurs. If the sunlight is not parallel then the optical path difference will change due to change in angle of incidence. Hence, the film will show different colors when viewed from different direction.

When a monochromatic light beam is incident on a transparent parallel thin film of uniform thickness t’ the rays reflected from the top and bottom of the film interfere to from interference pattern. These fringes are equally spaced alternate dark and bright bands.

The thin film interference is formed by division of amplitude of the ray which is partially reflected from the top and partially from the bottom of the film. Considered a transparent film of
thickness ‘t’ and refractive index ‘μ’

Let ray AB is partly reflected along BC and partly refracted along BM. At M part of it is reflected from lower surface of the film along MD and finally emerges along MK. These rays BC and DE interfere and interference fringes are produced. The intensity at any Point depends on the path difference between the interfering rays.

Let GM be the perpendicular from M on BD and HD be perpendicular from D on BC and Determination of path difference between the rays BM and MD. As shown in fig. 1

The geometric path difference between ray 1 and ray 2 = MF + FD – BH

Optical path difference = Δ

$latex \begin{array}{l}=\mu(BM+FD)-\mu BH\\\\=\mu\left(BM+FD\right)-BH\left(for\;air\;\mu=1\right)……….\left(1\right)\end{array}$

$latex \begin{array}{l}In\;the\;\triangle BMD,\angle BMG=\angle GMD=\angle r\\Further,\end{array}$

$latex \cos\;r=\frac{MG}{BM}\;\;\;\;\;\;;\;\;\;\;\;BM=\frac{MG}{\cos\;r}$

But MG=t

$latex BM=\frac t{\cos\;r}\;……….\left(2\right)$

Similarly

$latex MD=\frac t{\cos\;r}\;……….\left(3\right)$

Now, In ΔBHD

$latex \begin{array}{l}\sin\;i=\frac{BH}{BD}\\\\BH=BD\;\sin\;r\;\;……….\left(4\right)\\\\BD=BG+GD\;\;………\left(5\right)\end{array}$

But
Therefore, In ΔBGM

$latex \begin{array}{l}\tan\;r=\frac{BG}{MG}\\\\BG=MG\;\tan\;r\\\\BG=t\;\tan\;r\end{array}$

Similarly, GD=t tan r
Put the values of BG and GD in equation 5, we get

$latex \begin{array}{l}BD=t\;\tan\;r+t\;\tan\;r\\\\BD=2t\;\tan\;r\end{array}$

Therefore, equation 4 becomes,

$latex BH=2t\;\tan\;r\;\sin\;i\;\;……….\left(6\right)$

Put equation 2,3 and 6 in equation 1 we get

$latex \begin{array}{l}\triangle=\mu\left(\frac t{\cos\;r}+\frac t{\cos\;r}\right)-2t\;\tan\;r\;\sin\;i\\\\\triangle=\mu\left(\frac{2t}{\cos\;r}\right)-2t\frac{\sin\;r}{\cos\;r}\sin\;i\end{array}$

But

$latex \begin{array}{l}\mu=\frac{\sin\;i}{\sin\;r}\\\\\sin\;i=\;\mu\;\sin\;r\\\\\triangle=\mu\left(\frac{2t}{\cos\;r}\right)-\frac{2\mu t}{\cos\;r}\sin^2r\\\\\triangle=\frac{2\;\mu t}{\cos\;r}\left(1-\sin^2\;r\right)\\\\\triangle=\frac{2\;\mu t}{\cos\;r}\cos^2r\\\\\triangle=2\;\mu t\;\cos\;r\end{array}$

When light is reflected from the surface of an optically denser medium, a phase change π is introduced. Correspondingly, path difference λ/2 is introduced, therefore The effective path difference = ∆ = 2μ t cos r – λ/2

Condition for Brightness and Darkness:
1. When path difference ∆ = n λ, where n = 0,1,2,………
Constructive interference takes place and film appears bright. Therefore 2μ t cos r – λ/2 = n λ (for maximum intensity)
2μ t cos r = (2n +1) λ/2

2. When path difference ∆ = (2n+1) λ/2, where n = 0, 1, 2,
Destructive interference takes place and film appears dark. Therefore 2μ t cos r – λ/2 = (2n+1) λ/2
2μ t cos r = (n+1) λ

Where n is integer, therefore (n+1) can be written as n
Therefore, 2μt cos r = n λ (for minimum intensity)
The number n is called the order of interference.