Question

The figure shows two isotropic point sources of light $(S_{1}$ and $S_{2})$ that emit in phase at wavelength $500 nm$ and at the same amplitude. A detection point $P$ is shown on the $x$ -axis that extends through source $S_{1}$ . The phase difference $ϕ$ between the light arriving at point $P$ from the two sources to be measured as $P$ is moved along the $x$ -axis from $x = 0$ to $x = + \infty$ . The results out to $x_{s} = 10 \times 10^{- 7} m$ are given in figure on the way out to $+ \infty$ . What is the greatest value of $x$ at which the light arriving at $P$ from $S_{1}$ is exactly out of phase with the light arriving at $P$ from $S_{2}$ ?

Accepted Answer

Given:

Wavelength of wave is, $λ = 500 nm = 5 \times 10^{- 7} m$ .

Let the distance be $ϕ$ .

The distance, $S_{1} S_{2} = d$ .

The path difference between rays originating from $S_{1}$ and $S_{2}$ and arriving at points on the $x > 0$ axis and interfering destructively is,

$P . D . = (n + \frac{1}{2}) λ . . . (1)$

From the figure, path difference $(P . D .)$ between the waves interfering destructively is given by,

$\Rightarrow P . D . = {({x_{p}}^{2} + d^{2})}^{\frac{1}{2}} - x_{p} . . . (2)$

$\Rightarrow P . D . = x_{p} {(1 + \frac{d^{2}}{{x_{p}}^{2}})}^{\frac{1}{2}} - x_{p}$ .

Using binomial expansion and neglecting higher order terms, we get,

$\Rightarrow P . D . = x_{p} (1 + \frac{1}{2} \frac{d^{2}}{{x_{p}}^{2}}) - x_{p}$

$\Rightarrow P . D . = \frac{1}{2} \frac{d^{2}}{x_{p}} . . . (3)$

Now, using equation $(1)$ , equation $(3)$ becomes,

$\Rightarrow P . D . = \frac{1}{2} \frac{d^{2}}{x_{p}} = (n + \frac{1}{2}) λ . . . (4)$

$n = 0, 1, 2, . . .$

Now, from the figure, at $x_{p} = 0,$ the phase difference is, $ϕ = 6 π$ .

Then, $P . D . = \frac{λ}{2 π} ϕ = 3 λ . . . (5)$

From equation $(2)$ , path difference, $P . D . = d = 3 λ$ .

For the greatest value of $x_{p}$ so that light arriving at $P$ from $S_{1}$ is exactly out of phase, $n = 0$ .

Now, equation $(4)$ becomes,

$\Rightarrow \frac{1}{2} \frac{d^{2}}{x_{p}} = \frac{λ}{2}$

$\Rightarrow x_{p} = \frac{d^{2}}{λ} = \frac{{(3 λ)}^{2}}{λ}$

$\Rightarrow x_{p} = {(3)}^{2} \times 500 \times 10^{- 9}$

$\Rightarrow x_{p} = 4500 \times 10^{- 9} m$

$\Rightarrow x_{p} = 4.5 μm$ .

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