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Figure 35-26 shows two rays of light, of wavelength 600nm, that reflectfrom glass surfaces separated by 150nm. The rays are initially in phase.

(a) What is the path length difference of the rays?

(b) When they have cleared the reflection region, are the rays exactly in phase, exactly out of phase, or in some intermediate state?

Short Answer

Expert verified

(a) The path length difference between the two waves is 300nm.

(b) When the rays have cleared the region, they are exactly out of phase.

Step by step solution

01

Given data: 

The wavelength of the two rays is λ=600nm
.

The thickness of the glass surface is d=300nm.

02

Relation between phase difference and path difference:

The phase difference between two waves of a wavelength λ having the path difference x isϕ=2πλx

03

(a) Determining the path difference of the rays:

The upper wave travels a distance twice the thickness greater than the lower wave. Thus, the path difference isx=2×150nm=300

Therefore, the path difference is 300nm.

04

Determining the phase difference of the rays:

From equation (1), the phase difference between the two waves is,ϕ=2π600nm×300nm=2π2=π

Hence, the phase difference is π
.

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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?

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