Visible light of frequency f = 5 x 10¹4 Hz is beamed directly at a metal reflector. Light is traveling in oil, where its speed is 1.5 x 10 m/s. a) Calculate the wavelength of the visible light. b) Consider A = 6 x 10 m. The spacing Ar between successive antinodes in the resulting standing-wave pattern is given by: 6 x 10-7 2 m. c) Az Calculate Δz. 772 6 x 10-7 3 6 x 10-7
Visible light of frequency f = 5 x 10¹4 Hz is beamed directly at a metal reflector. Light is traveling in oil, where its speed is 1.5 x 10 m/s. a) Calculate the wavelength of the visible light. b) Consider A = 6 x 10 m. The spacing Ar between successive antinodes in the resulting standing-wave pattern is given by: 6 x 10-7 2 m. c) Az Calculate Δz. 772 6 x 10-7 3 6 x 10-7
Related questions
Question

Transcribed Image Text:Problem 2
This problem is composed of two independent parts.
For all parts of this problem, fill in the empty boxes with your answer (YOUR ANSWER MUST BE ONLY
A NUMBER; DO NOT WRITE UNITS; DO NOT WRITE LETTERS).
Part 1
Visible light of frequency f = 5 x 10¹4 Hz is beamed directly at a metal reflector. Light is traveling in oil,
where its speed is 1.5 x 108 m/s.
a)
Calculate the wavelength of the visible light.
A =
b)
Consider A = 6 x 107 m. The spacing Ar between successive antinodes in the resulting
standing-wave pattern is given by:
6 x 10-7
2
c)
Ar
E₁
m
=
b)
E₂ =
x =
Calculate Δ.r.
Part 2
A stal...ing wave is given by E= 400 sin(7.14 x) cos ( 11.42 t). Two waves E₁ and E, can be
superimposed to generate this standing wave.
772
6 x 10-7
3
Determine the wave E, as per the below:
sin(
t)
Determine the wave En as per the below:
sin(
t)
m
6 x 10-7
4
c)
The wavelength of this wave is 0.87 m. For x ≥ 0, what is the location of the antinode
having the smallest value of a?
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 5 steps with 5 images
