Question

In Fig. 35-35, two light rays go through different paths by reflecting from the various flat surfaces shown.The light waves have a wavelength of 420.0 nm and are initially in phase. What are the (a) smallest and (b) second smallest value of distance L that will put the waves exactly out of phase as they emerge from the region?

Short answer

a. The least possible length that will result in destructive interference is 52.50 nm.

b. The second least possible length that will result in destructive interference is 157.5 nm.

Step-by-step solution

Definition of reflection.

An electromagnetic wave incident over an obstacle with a certain angle reflects that wave maintaining the same angle for a flat surface. This reflected wave is called a reflection of the incident wave.

Determination of the smallest value of distance that will put the waves out of phase.

a.

In figure 35.35, we can conclude that ray 1 travels an extra distance of $4L$more than ray 2.

Therefore to get the least possible length that will result in destructive interference, it is required to set the extra distance equated to half of the wavelength.

$$\begin{gathered} 4L=\frac{\lambda }{2} \\ L=\frac{\lambda }{8} \end{gathered}$$.

Here it is given that the wavelength $\lambda$ is 420 nm.

Therefore,

$$\begin{gathered} L=\frac{420\times {10}^{-9}}{8} \\ =52.50\text{nm} \end{gathered}$$.

Determination of the second smallest value of distance that will put the waves out of phase.

b.

In figure 35.35, we can conclude that ray 1 travels an extra distance of $4L$more than ray 2.

Therefore to get the second least possible length that will result in destructive interference, it is required to set the extra distance equated $\frac{3}{2}$of the wavelength.

$$\begin{gathered} 4L=\frac{3}{2}\lambda \\ L=\frac{3}{8}\lambda \\ L=\frac{3}{8}\times \left(420\text{nm}\right) \\ L=157.5\text{nm} \end{gathered}$$.