Theory of Holograpgy

Theory of Holography:

If the object is a point scatter, then the object wave would just be \(\frac{A}{r} cos(kr - \omega t + \phi) \) where, r represents the distance of the point of observation from the point scatter and A represents a constant ; k = \(\frac{2 \pi}{\lambda}\). An object can be thought of as being made up of a large number of points and the composite wave reflected by the light would be vectorial sum of these. The fundamental problem in holography is the recording of this object wave, in particular the phase distribution associated with it.

$$O(x, y)= a(x, y) cos[\phi(x, y)- \omega t]\dotsm(1)$$

This equation present the object wave. These waves from the point scatters superpose in the plane of photographic plate which is assumed to be zero for z-axis. Now, let us consider a plane reference wave and assumed that it is travelling in the x-z plane inclined at an angle \(\theta\) with the z-direction. The field associated with this plane would be given by,

$$r(x, y, z)=A cos[k.r-\omega t]$$

$$=A cos[k(x sin\theta+z cos\theta)-\omega t]$$

$$=Acos [kx sin\theta+kz cos\theta-\omega t]\dotsm(2)$$

If r(x, y) represents the field at the plane z = 0, due to this reference wave, then one can see that.

$$r(x, y)=A cos [kx sin\theta-\omega t]$$

$$=A cos\biggl[\frac{2 \pi}{\lambda} x sin\theta-\omega t\biggr]$$

$$=A cos\biggl[2\pi \biggl(\frac{sin\theta}{\lambda}\biggr) x-\omega t\biggr]$$

$$=A cos [2\pi \alpha x-\omega t]\dotsm(3)$$

Where, \(\frac{sin\theta}{\lambda}=\alpha\). It is the spatial frequency. Equation (3) represents the field due to a plane wave inclined at an angle \(\theta\) with the z-axis and the phase varies linearly with x. There is no y-dependence because the plane wave has been assumed to have its propagation vector in the x-z plane. Thus, the total field at the photographic plate would be given by

$$u(x, y , t)=a(x, y)cos[\phi(x, y)-\omega t]+A cos[2 \pi \alpha x-\omega t]\dotsm(4)$$

The photographic plate respond only to the intensity which would be proportional to the time average of [ u(x, y, t)\(]^2\). Thus, the intensity pattern recorded by the photographic plate would be

$$I(x, y)=<u^2(x, y, t)>$$

$$=<[a(x, y) cos{\phi(x, y)-\omega t}+A cos(2\pi \alpha x-\omega t)]^2>\dotsm(5)$$

Where the angular brackets denote the time averaging. Thus,

$$I(x, y)=a^2(x, y)<cos^2{\phi(x, y)-\omega t}>+A^2<cos^2(2\pi \alpha x-\omega t)>+2a(x, y)A<cos{\phi(x, y)-\omega t} cos(2\pi \alpha x-\omega t)>\dotsm(6)$$

Since, \(<cos^2[\phi(x, y)-\omega t]>=\frac{1}{2}=<cos^2[2\pi \alpha x-\omega t)]>\dotsm(7)\)

And <cos[\phi(x, y)-\omega t]cos(2\pi \alpha x-\omega t)>

$$=\frac{1}{2}<cos[\phi(x, y)+2\pi \alpha x-\omega t]>+\frac{1}{2}<cos[\phi(x, y)-2\pi \alpha x]>$$

$$=\frac{1}{2}cos[\phi(x, y)-2\pi \alpha x]\dotsm(8)$$

$$\biggl[\therefore cos A.cos B=\frac{1}{2}cos(A + B)+\frac{1}{2}cos(A-B)\biggr]$$

Thus, equation (6) becomes $$I(x, y)=\frac{1}{2}a^2(x, y)+\frac{1}{2}A^2+Aa(x, y)cos[\phi(x, y)-2\pi \alpha x]\dotsm(9)$$

From the above relation, it is obvious that the phase information of the object wave, which is contained in f(x, y) is recorded in the intensity pattern.

When the photographic plate ( which has recorded the above intensity pattern ) is developed, one obtains a hologram. The transmittance of the hologram, i.e. the ratio of the transmitted field to the incident field, depends on I(x, y). By a suitable developing process one can obtain a condition under which the amplitude transmittance would be linearly related to I(x, y). Thus, in such a case if R(x, y) represents the field of the reconstruction wave at the hologram plane, then the transmitted field would be given by V(x, y) = KR(x, y) I(x, y)

$$=K\biggl[\frac{1}{2}a^2(x, y)+\frac{1}{2}A^2\biggr]R(x, y)+K Aa(x, y) cos[\phi(x, y)-2\pi \alpha x]\dotsm(10)$$

Where, K is constant. We consider the case when the reconstruction wave is identical to the reference wave r(x, y). In such case, we would obtain (omitting the constant K).

$$v(x, y)=\biggl[\frac{1}{2}a^2(x, y)+\frac{1}{2}A^2\biggr]Acos(2\pi \alpha x-\omega t)+A^2a(x, y)cos[2\pi \alpha x-\omega t]cos[\phi(x, y)-2\pi \alpha x]$$

$$=\biggl[\frac{1}{2}a^2(x, y)+\frac{1}{2}A^2\biggr]Acos(2\pi \alpha x-\omega t)+\frac{1}{2}A^2a(x, y)cos[\phi(x, y)-\omega t]+\frac{1}{2}A^2a(x, y)cos[4\pi\alpha x-\phi(x, y)-\omega t]\dotsm(11)$$

Equation (11) gives the transmitted field in the plane z = 0. We consider each of the three terms separately. The first term is nothing but the reconstruction wave itself whose amplitude is modulated due to the presence of the term \(a^2(x, y)\). This part of the total field is travelling in the direction of the reconstructed wave. The second term is identical to the R.H.S. of the equation (1) and hence represents the original object wave. This gives rise to a virtual image. Thus the effect of viewing this wave is the same as viewing the object itself. The reconstructed object wave is travelling in the same direction as the original object wave.

To study the last term, we first observe that In addition to the term 4\(\pi \alpha x\), the phase term \(\phi(x, y)\) carries a negative sign. The negative sign indicates that the wave has a curvature opposite to that of the object wave. Thus, if the object wave is a diverging spherical wave, then the last term represents a converging spherical wave. Thus, in contrast to the second term, this wave forms a real image of the object which can be photographed by simply placing a film.

References:

Adhikari, P.B, Daya Nidhi Chhatkuli and Iswar Prasad Koirala. A Textbook of Physics. Vol. II. Kathmandu: Sukunda Pustak Bhawan, 2012.

Jenkins, F.A and H.E White. Fundamental of optics. New York (USA): McGraw-Hill Book Co, 1976.

wood, R.W. Physical Optics. New York (USA): Dover Publication , 1934.