An electromagnetic wave of intensity 50 Wm–2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by:

( 1 \over \sqrt n ,1 \over \sqrt n )

Electromagnetic Waves PYQ — JEE Main 2019 · 11 Jan · Shift 1 · Q42 with Solution

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Correct answer: C

Intensity of an electromagnetic wave

For a plane electromagnetic wave, the average intensity is

$I = \frac{1}{2} E_0 H_0$

Also, using wave speed $v$ ,

$I = \frac{1}{2}\,\varepsilon E_0^2 v = \frac{1}{2}\,\frac{B_0^2}{\mu}v$

For a non-magnetic medium, we take $\mu = \mu_0$ and refractive index

$n = \frac{c}{v} \implies v = \frac{c}{n}$

and

$\varepsilon = n^2 \varepsilon_0$

Given: no loss on entering the medium

So intensity remains unchanged:

$I_1 = I_2$

Electric field ratio

Using

$I = \frac{1}{2}\varepsilon E_0^2 v$

In air/vacuum:

$I_1 = \frac{1}{2}\varepsilon_0 E_1^2 c$

In medium:

= \frac{1}{2} n \varepsilon_0 E_2^2 c$$ Since $I_1 = I_2$, $$\frac{1}{2}\varepsilon_0 E_1^2 c = \frac{1}{2} n \varepsilon_0 E_2^2 c$$ $$E_1^2 = n E_2^2$$ $$\frac{E_1}{E_2} = \sqrt{n}$$ So the ratio of electric field magnitudes **before : after** is $$\sqrt{n}$$ 4. **Magnetic field ratio** For an electromagnetic wave, $$E = vB$$ Before entering (vacuum/air): $$E_1 = cB_1$$ After entering medium: $$E_2 = \frac{c}{n} B_2$$ Thus, $$B_1 = \frac{E_1}{c}, \qquad B_2 = \frac{nE_2}{c}$$ Therefore, $$\frac{B_1}{B_2} = \frac{E_1/c}{nE_2/c} = \frac{E_1}{nE_2}$$ Using $\dfrac{E_1}{E_2} = \sqrt{n}$, $$\frac{B_1}{B_2} = \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}$$ So the ratio of magnetic field magnitudes **before : after** is $$\frac{1}{\sqrt{n}}$$ 5. **Matching with options** Thus the required pair is $$\left(\sqrt{n}, \frac{1}{\sqrt{n}}\right)$$ This matches **Option C**.

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View written solutionFree

Correct answer: C

Intensity of an electromagnetic wave

For a plane electromagnetic wave, the average intensity is

$I = \frac{1}{2} E_0 H_0$

Also, using wave speed $v$ ,

$I = \frac{1}{2}\,\varepsilon E_0^2 v = \frac{1}{2}\,\frac{B_0^2}{\mu}v$

For a non-magnetic medium, we take $\mu = \mu_0$ and refractive index

$n = \frac{c}{v} \implies v = \frac{c}{n}$

and

$\varepsilon = n^2 \varepsilon_0$

Given: no loss on entering the medium

So intensity remains unchanged:

$I_1 = I_2$

Electric field ratio

Using

$I = \frac{1}{2}\varepsilon E_0^2 v$

In air/vacuum:

$I_1 = \frac{1}{2}\varepsilon_0 E_1^2 c$

In medium:

= \frac{1}{2} n \varepsilon_0 E_2^2 c$$ Since $I_1 = I_2$, $$\frac{1}{2}\varepsilon_0 E_1^2 c = \frac{1}{2} n \varepsilon_0 E_2^2 c$$ $$E_1^2 = n E_2^2$$ $$\frac{E_1}{E_2} = \sqrt{n}$$ So the ratio of electric field magnitudes **before : after** is $$\sqrt{n}$$ 4. **Magnetic field ratio** For an electromagnetic wave, $$E = vB$$ Before entering (vacuum/air): $$E_1 = cB_1$$ After entering medium: $$E_2 = \frac{c}{n} B_2$$ Thus, $$B_1 = \frac{E_1}{c}, \qquad B_2 = \frac{nE_2}{c}$$ Therefore, $$\frac{B_1}{B_2} = \frac{E_1/c}{nE_2/c} = \frac{E_1}{nE_2}$$ Using $\dfrac{E_1}{E_2} = \sqrt{n}$, $$\frac{B_1}{B_2} = \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}$$ So the ratio of magnetic field magnitudes **before : after** is $$\frac{1}{\sqrt{n}}$$ 5. **Matching with options** Thus the required pair is $$\left(\sqrt{n}, \frac{1}{\sqrt{n}}\right)$$ This matches **Option C**.

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