Complete Question
A spherical wave with a wavelength of 2.0 mm is emitted from the origin. At one instant of time, the phase at r_1 = 4.0 mm is π rad. At that instant, what is the phase at r_2 = 3.5 mm ? Express your answer to two significant figures and include the appropriate units.
Answer:
The phase at the second point is [tex]\phi _2 = 1.57 \ rad[/tex]
Explanation:
From the question we are told that
The wavelength of the spherical wave is [tex]\lambda = 2.0 \ mm = \frac{2}{1000} = 0.002 \ m[/tex]
The first radius is [tex]r_1 = 4.0 \ mm = \frac{4}{1000} = 0.004 \ m[/tex]
The phase at that instant is [tex]\phi _1 = \pi \ rad[/tex]
The second radius is [tex]r_2 = 3.5 \ mm = \frac{3.5}{1000} = 0.0035 \ m[/tex]
Generally the phase difference is mathematically represented as
[tex]\Delta \phi = \phi _2 - \phi _1[/tex]
this can also be expressed as
[tex]\Delta \phi = \frac{2 \pi }{\lambda } (r_2 - r_1 )[/tex]
So we have that
[tex]\phi _2 - \phi _1 = \frac{2 \pi }{\lambda } (r_2 - r_1 )[/tex]
substituting values
[tex]\phi _2 - \pi = \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 )[/tex]
[tex]\phi _2 = \frac{2 \pi }{0.002 } ( 0.0035 - 0.004 ) + 3.142[/tex]
[tex]\phi _2 = 1.57 \ rad[/tex]
