Question

What is the path difference for the first, second and third order maxima of wavelengths $4000A°$ and $7000A°$?

Answer 1

The principle to be applied here, to solve this problem, is the principle of interference of light waves. There are two cases of interference of light namely, constructive interference and destructive interference. The constructive interference is related to the maxima and hence, we have to consider the concept of constructive interference here.

Complete step by step answer:
The interference of light waves occurs when two coherent waves superimpose on one another. There are two results that happen while interference of light occurs:
i) Constructive interference: When the two waves are in phase, the amplitude of the waves add together resulting in a greater amplitude.
ii) Destructive interference: When the two waves meet in opposite phases, the amplitudes of the waves cancel out each other and the resulting amplitude is equal to the difference between the two amplitudes.
The criteria for defining interference is based on the path difference between the waves, which is the difference between the distances travelled by the two waves to a certain point. The criteria is given as follows –
If the path difference between the waves is equal to even multiples of $\dfrac{\lambda }{2}$, constructive interference occurs.
$\Rightarrow D = 2n\dfrac{\lambda }{2} = n\lambda$
If the path difference between the waves is equal to odd multiples of $\dfrac{\lambda }{2}$ , constructive interference occurs.
$\Rightarrow D = \left( {2n + 1} \right)\dfrac{\lambda }{2}$
The maximum occurs when there is constructive interference and the number n is a natural number $\left( {n = 1,2,3..} \right)$ that represents the subsequent order of maxima.
i) For the wavelength, $\lambda = 4000A°$
The first maximum occurs at $n = 1$
Thus, path difference
$D = n\lambda = \lambda = 4000A°$
The second maximum occurs at $n = 2$
Thus, path difference
$\Rightarrow D = n\lambda = 2\lambda = 2 \times 4000 = 8000A°$
The third maximum occurs at $n = 3$
Thus, path difference
$\Rightarrow D = n\lambda = 3\lambda = 3 \times 4000 = 12000A°$
ii) For the wavelength, $\lambda = 7000A°$
The first maximum occurs at $n = 1$
Thus, path difference
$\Rightarrow D = n\lambda = \lambda = 7000A°$
The second maximum occurs at $n = 2$
Thus, path difference
$\Rightarrow D = n\lambda = 2\lambda = 2 \times 7000 = 14000A°$
The third maximum occurs at $n = 3$
Thus, path difference
$\Rightarrow D = n\lambda = 3\lambda = 3 \times 7000 = 21000A°$

Note: Interference can also, be explained in terms of phase difference as follows:
1. If the phase difference between the waves is equal to even multiples of ${90^ \circ }$, constructive interference occurs.
2. If the phase difference between the waves is equal to odd multiples of ${90^ \circ }$, destructive interference occurs.