Chapter 35: Q25P (page 1075)

In Fig. 35-40, two isotropic point sources of light (S₁ and S₂) are separated by distance $2.70 μ m$ along a y axis and emit in phase at wavelength 900 nm and at the same amplitude. A light detector is located at point P at coordinate $x_{P}$ on the x axis. What is the greatest value of $x_{P}$ at which the detected light is minimum due to destructive interference?

Short Answer

Expert verified

The maximum value of point P position for destructive interference is $7.8 μ m$ .

Step by step solution

Identification of given data

The separation between both sources is $d = 2.70 mm$

The wavelength of the light is $λ = 900 nm$

The phase difference of fringe pattern varies with the path difference. For minimum phase difference path difference should be minimum and vice versa.

Determination of maximum value for position of point P for destructive interference

The maximum value of point P position for destructive interference is given as:

$x_{P} = \frac{d^{2}}{(2 m + 1) λ} - \frac{(2 m + 1) λ}{4}$

Here,m is the order of fringe for minimum intensities at various positions and its minimum value for minimum possible path difference is zero.

Substitute all the values in equation.

$\begin{array}{rcl} x_{P} & = & \frac{{[(2.70 μ m) (\frac{10^{- 6} m}{1 μ m})]}^{2}}{(2 (0) + 1) [(900 nm) (\frac{10^{- 9} m}{1 nm})]} - \frac{(2 (0) + 1) [(900 nm) (\frac{10^{- 9} m}{1 nm})]}{4} \\ x_{P} & = & 8.1 \times 10^{- 6} m - 0.225 \times 10^{- 6} m \\ x_{P} & = & (7.8 \times 10^{- 6} m) (\frac{1 μ m}{10^{- 6} m}) \\ x_{P} & = & 7.8 μ m \end{array}$