Answer to Question #105801 in Optics for Ambika

Question #105801
Light waves are propagating in vacuum. Derive the wave equation for the
associated magnetic field vector. On the basis of this equation, calculate the speed
of light.
1
Expert's answer
2020-05-05T18:50:45-0400

Let the electric field and the magnetic field vector is along the y axis and along the z axis.

The linearly polarized plane wave is traveling along the x axis and let the speed of light is c.

"E\n=Eo\u200b(x,t)j^\u200b" and "B\n=Bo\u200b(x,t)k"

where x is the displacement along the x axis, t is the time.

As per the maxwell's equation for the space

"\u2207.E=0" and "\u2207.B=0"

"\u2207\u00d7E=\u2212\\dfrac{\u2202t}{\u2202B}" and "\u2207\u00d7B=\\mu_{o}\u200b\\varepsilon_{o}\u200b\\dfrac{\u2202t}{\u2202E}"

now,



Now, equating the magnitudes of the faradays law

"\\dfrac{\u2202x}{\u2202E}\u200b=\u2212\\dfrac{\u2202t}{\u2202B}"

now taking the partial derivative

"\\dfrac{\u2202x^2}{\u2202^2E}\u200b=\u2212\\dfrac{\u2202t^2}{\u2202^2B}"



Similarly



Now from the the above

we know that the general equation of the wave travailing along the x axis

"\\dfrac{\u2202\u03c8^2}{\u2202x}\u200b=\\dfrac{\u2202\u03c8^2}{\u03bd^2\u2202x}"

From the second derivative of electric and magnetic field

"\\mu_{o}\u200b\\varepsilon_{o}\u200b=\\dfrac{1}{c^2}"

"c=\\dfrac{1}{\\mu_{o}\u200b\\varepsilon_{o}}\u200b\u200b=\\dfrac{1}{8.85\u00d710^{\u221212}\u00d74\u03c0\u00d710^{\u221271}\u200b}m\/sec"

"c=2.97\u00d710^8m\/sec"


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