Question

Figure 35-30 shows three situations in which two rays of sunlight penetrate slightly into and then scatter out of lunar soil. Assume that the rays are initially in phase. In which situation are the associated waves most likely to end up in phase? (Just as the Moon becomes full, its brightness suddenly peaks, becoming 25% greater than its brightness on the nights before and after, because at full Moon we intercept light waves that are scattered by lunar soil back toward the Sun and undergo constructive interference at our eyes. Before astronauts first landed on the Moon, NASA was concerned that backscatter of sunlight from the soil might blind the lunar astronauts if they did not have proper viewing shields on their helmets.)

Short answer

The situations in which the associated waves are most likely to end up in phase are (a) and (c).

Step-by-step solution

Given information 

In cases (a), (b), and (c), the light rays are initially in phase.

Phase difference

The phase difference between two light waves traveling through mediums relies upon the indexes of refraction of mediums, the wavelength of light waves, and the path length of each wave.

The phase difference formula between two waves traveling in the same direction is given by,

$N_2-N_1=\frac{L_2{n}_2}{\lambda }-\frac{L_1{n}_1}{\lambda }$

…(1)

Here, N₁ is the phase of ray 1, N₂ is the phase of ray 2, L is the path length of ray 1, L₂is the path length of ray 2, n₁ is the refractive index of medium 1, n₂is the refractive index of medium 2 and$\lambda$ is the wavelength of the light ray.

Step 3(a): Wave phase difference

From the figure, the path lengths and direction of both rays are the same and the medium is also the same. So, putting L₂= L₁ = L and n₂ = n₁ = n in equation (1),

$\begin{array}{l}{N}_2-N_1=\frac{Ln}{\lambda }-\frac{Ln}{\lambda }\\ N_2-N_1=0\end{array}$

Hence, the waves are likely to end up in phase.

Step 4(b): Wave phase difference

From the figure, the value of path length is different while the direction of both rays and the medium is the same. So, putting $L_2\ne L_1$and n₂ = n₁ = n in equation (1),

$\begin{array}{l}{N}_2-N_1=\frac{L_{2}n}{\lambda }-\frac{L_{1}n}{\lambda }\\ N_2-N_1=\frac{n}{\lambda }\left(L_2-L_1\right)\end{array}$

Hence, the waves are not likely to end up in phase.

Step 5(c): Wave phase difference 

From the figure, the value of the path length of both rays is the same while the direction of both rays is opposite to each other in the same medium. So, putting L₂= L₁ = L and n₂= n₁= n in equation (1),

$\begin{array}{l}{N}_2-N_1=\frac{L_{2}n}{\lambda }-\left(-\frac{L_{1}n}{\lambda }\right)\\ N_2-N_1=\frac{n}{\lambda }\left(L+L\right)\\ N_2-N_1=\frac{2Ln}{\lambda }\end{array}$

Hence, the waves are likely to end up in phase.