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An electric field is essentially a force field created around a charged particle.

A magnetic field is a magnetic field generated around a permanent magnet or a moving charged object.

As you know, visible light is part of the electromagnetic spectrum. The velocity of the electromagnetic wave is given by the following equation.

\(c = \nu \lambda \)

Also, we know that the visible light has wavelengths between\(380\;{\rm{nm}}\)and\(760\;{\rm{nm}}\). As we realize the wavelength of the wavelength of electromagnetic wave in a medium is given by

\({\lambda _n} = \dfrac{\lambda }{n}\)

Where,\(\lambda \)is the wavelength of the electromagnetic wave in vacuum and\(n\)is the redfraction index of the given medium. As we know the thickness of any material is given by

\(t = \dfrac{\lambda }{n}\)

As we know, Young's double-slit experiment has unusual results. As we know, light must interact with something as small as something. B. Narrow slot used by Young. To get the constructive interference of the double slit, the difference in optical path length must be an integral multiple of the wavelength.

\(d\sin (\theta ) = m\lambda \)

To get the destructive interference of the double slit, the difference in optical path length must be a half-integer multiple of the wavelength.

\(d\sin (\theta ) = \left( {m + \dfrac{1}{2}} \right)\lambda \)

Where,\({\rm{d}}\)is the distance between the slits,\(\theta \)is the angle from the original direction of the beam and\(m\)is the order of the interference.

As you know, light is polarized as it passes through a polarizing filter and other polarizing materials. Follow the Malus` law:

\(I = {I_0}{\cos ^2}(\theta )\)

Also, Brewster's law assigns that

\(\tan \left( {{\theta _b}} \right) = \dfrac{{{n_2}}}{{{n_1}}}\).

Solve for first case:

As we know the relation between the electric field and the magnetic field is

\({I_1} = cB_1^2\)

Solve for second case:

As we know the relation between the electric field and the magnetic field is

\({I_2} = cB_2^2\)

Divide both equations together, then we get

\(\begin{aligned}\dfrac{{{I_1}}}{{{I_2}}} &= \dfrac{{cB_1^2}}{{cB_2^2}}\\ &= \dfrac{{B_1^2}}{{B_2^2}}\dfrac{{{I_1}}}{{\dfrac{1}{2}{I_1}}}\\ &= 2\dfrac{{B_1^2}}{{B_2^2}}\dfrac{{B_1^2}}{{B_2^2}}\end{aligned}\)
Rearrange and solve for the magnetic field in the second case \({B_2}\):

\({B_2} = \sqrt {\dfrac{1}{2}} {B_1}\)

So, the electric field and the magnetic field will reduce by \(\sqrt {\dfrac{1}{2}} \) of its original value.