# Interference of light waves question

We were recently asked to solve a question in class which goes as follows:

In a modified Young's double slit experiment, a monochromatic uniform and parallel light beam of wavelength $6000$ angstrom and intensity $(10/\pi)\,\mathrm{W\,m^{-2}}$ is incident normally on two circular apertures A and B of radii $10^{-3}\,\mathrm m\;$ and $2\times10^{-3}\,\mathrm m\;$ respectively. A perfectly transparent film of thickness $2000$ angstrom and refractive index $1.5$ for the given wavelength of light, is placed in front of aperture A as shown in the figure. Calculate the power received at the focal spot F of the lens. The lens is placed symmetrically with respect to the apertures. Assume that $10\%$ of the power received by each aperture goes in the original direction and is brought to the focal spot.

We have found that the power transmitted by the aperture A is $P_A=10^{-6}\,\text W$ and by the aperture B is $P_B=4\times10^{-6}\,\text W$, and the phase difference between the two beams from each aperture to be $\delta=\pi/3$.

Now, our teacher says that assuming the power of each beam at the focal spot to be proportional to the intensity of the respective beam, we can convert the formula, $$I_{\text{resultant}}=I_A + I_B + 2\sqrt{I_AI_B}\cos(\delta)$$ to, $$P_{\text{resultant}}=P_A + P_B + 2\sqrt{P_AP_B}\cos(\delta)$$

But this results in an answer of $7\times10^{-6}\,\text W$, which implies that an extra power of 2 microwatts is appearing from thin air. So, is the answer and the approach correct? If the answer is correct, then where does the extra power come from? If not, what is the correct approach? Also, then why is the formula for resultant intensity invalid?

Note: I have written intensity everywhere to mean irradiation, because for now, we call it intensity where we learn about it.

Thank you.

Well spotted.

The average intensity of the whole fringe system must be $1+4=5\;\mu$W.

However there will be places where the intensity is a maximum, $1+4+2\sqrt{1\times4} = 9\;\mu$W and places where the intensity is a minimum, $1+4-2\sqrt{1\times4} = 1\;\mu$W.

Your chosen position is one which is nearer a maximum than a minimum.

Later

Here is a superb photograph showing the diffraction patterns from two circular holes which modulate the two hole interference pattern.
Near the centre it can be seen that the interference fringes are not straight.
The maxima of the diffraction patterns are overexposed so that the interference fringes can be seen.

• Thank you for the answer! Just one more question out of curiosity, will the interference pattern be in the form of circular rings of brightness and darkness? – FreezingFire Apr 12 '16 at 8:09
• No. The fringes will be curved although usually they appear to be straight. I have found a nice picture of the pattern from two holes here (Figure 5) . home.myfairpoint.net/vzeeg3o2/id5.html – Farcher Apr 12 '16 at 8:57
• But, there is a lens after the two apertures. Wouldn't it affect the interference pattern? The lens is supposed to converge the two parallel beams of light at the focal spot... – FreezingFire Apr 12 '16 at 9:15
• From the two holes there are rays going in all directions. The rays which are parallel to the principal axis of the lens will meet at the focal point of the lens. Other rays which are parallel will meet in the focal plane of the lens. In this way you will not just see a spot of light rather you will see the whole diffraction/interference pattern.as above. – Farcher Apr 12 '16 at 9:38