## Position of First–Order Maxima When Two Coherent Rays Interfere

### Task number: 1699

The distance between two coherent light sources is 0.45 mm. Determine the distance between the zero-order maximum and the first red stripe of a wavelength 700 nm, provided that the screen is at a distance 0.5 m from the sources.

#### Hint 1

What is the path difference of two rays which are emitted from two sources and incident on the same point of the screen?

#### Hint 2

What is the path difference of two rays in order to create an interference maximum?

#### Hint 3

What is the relationship between the distance of the first red maximum Δ

*x*,which occurs on the screen, and the angle of deflection*α*?#### Given values

*b*= 0.45 mm = 4.5·10^{−4}mDistance between two light sources *λ*= 700 nm = 7.0·10^{−7}mWavelength of light *k*= 1Order of maximum *d*= 0.5 mDistance between the sources and the screen Δ *x*= ? (mm)Distance between the first-order red maximum and the zero-order maximum #### Solution

The light propagates from its source in all directions. The light from both sources is incident on the screen, and an interference pattern is created. On the screen the interference maxima and minima are formed; their position depends on the path difference of the two rays emitted from the sources.

Suppose that the screen is far from the sources, so that the rays coming to a point A of the screen are approximately parallel.

From the triangle in the picture we obtain a relationship for the path difference Δ

\[\mathrm{\Delta}l = \,b\sin\,\alpha,\]*l*of two adjacent rays::where

*b*is the distance between the sources and*α*is the angle between the rays and the central axis.In the point A the rays interfere and create a stripe, a maximum. Thus the path difference Δ

\[\mathrm{\Delta}l = k \lambda,\]*l*must satisfy the condition:where k = 1. From these two equations we evaluate sin

\[\sin\,\alpha=\,\frac{\lambda}{b}.\]*α*:We draw a picture where we denote

*d*as the distance between the sources and the screen,*α*as deviation angle, Δ*x*as the distance between the first-order red maximum and the zero-order maximum.We obtain:

\[\mathrm{tg}\,\,\alpha=\, \frac{\mathrm{\Delta}x}{d} \quad\Rightarrow\quad \mathrm{\Delta}x=\,d\,\mathrm{tg}\,\,\alpha,\]where

\[\sin\,\alpha=\,\frac{\lambda}{b}=\,\frac{7.0{\cdot}10^{−7}}{4.5{\cdot}10^{−4}}\,\dot=\,0.00156 \quad\Rightarrow\quad \alpha\,\dot=\,0.00156,\]*α*is calculated from the relation:and we substitute the values given in the assignment.

\[\mathrm{\Delta}x=\,d\,\mathrm{tg}\,\,\alpha=\,0.5\,\cdot(\mathrm{tg}\,\,0.00156)\,\mathrm{m}\,\dot=\, 8.0{\cdot}10^{−4}\,\mathrm{m}\,\dot= \,0.80\,\mathrm{mm}. \]**Note to the solution:**To simplify the calculation we can take into account that we deal with very small angles, therefore

\[\sin\,\alpha\ \dot{=}\ \, \alpha\ \dot{=}\ \mathrm{tg}\,\,\alpha,\]where the middle equality is valid if the angle α is in radians.

Now we solve this task using this finding.

We substitute the relation:

\[\sin\,\alpha=\,\frac{\lambda}{b},\]into the equation for Δ

\[\mathrm{\Delta}x=\,d\,\mathrm{tg}\,\,\alpha=\,d\,\frac{\lambda}{b}.\]*x*:We substitute given values.

\[\mathrm{\Delta}x=\,d\,\frac{\lambda}{b}=\,0.5\cdot\,\frac{7.0{\cdot}10^{−7}}{4.5{\cdot}10^{−4}}\,\mathrm{m}\,\dot=\,7.8{\cdot}10^{−4}\,\mathrm{m}\,\dot=\,0.78\,\mathrm{mm}.\]Thus we can see that the task can be solved with this neglect. This result is not very different from the first result.

#### Answer

The first red stripe appears approximately 0.8 mm from the zero-order maximum.

#### Link to experiment

If you would like to see the interference pattern created on the screen, look at the Double-Slit Experiment.