A monochromatic beam of light has a frequency ν = 3/2π × 10¹² Hz and is propagating along the direction (i + j)/(√(2)). It is polarized along the k direction. The acceptable form for the magnetic field is: (a) (b) (c) (d)

Answer: (a). ω = 2πν = 3 × 10¹² rad/s and k = (ω)/(c) = 3 × 10¹²3 × 10⁸ = 10⁴ m ⁻¹. Wave along +n: phase (10⁴ n - 3 × 10¹² t) with n = (i + j)/(√(2)). B = n × E = (i + j)/(√(2)) × k = (i - j)/(√(2)), and B₀ = E₀/c. So the answer is (a) E₀/c(i - j)/(√(2))cos [10⁴(i + j)/(√(2)) - (3 × 10¹²)t].

A monochromatic beam of light has a frequency ν = 3/2π × 10¹² Hz and… | JEE Main 2018 (15 Apr, Shift 1)

Q72JEE Main 2018 · 15 APR · SHIFT 1Difficult

Nature of EM Waves & Wave Equation

A monochromatic beam of light has a frequency $\nu = \frac{3}{2\pi} \times 10^{12}\text{ Hz}$ and is propagating along the direction $\frac{\hat{i} + \hat{j}}{\sqrt{2}}.$

It is polarized along the $\hat{k}$ direction. The acceptable form for the magnetic field is:

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Answer: (a)

$\omega = 2\pi\nu = 3 \times 10^{12}$ rad/s and $k = \dfrac{\omega}{c} = \dfrac{3 \times 10^{12}}{3 \times 10^{8}} = 10^{4}$ m $^{-1}$ .
Wave along $+\hat{n}$ : phase $\left(10^{4}\,\hat{n}\cdot\vec{r} - 3 \times 10^{12} t\right)$ with $\hat{n} = \dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$ .
$\hat{B} = \hat{n} \times \hat{E} = \dfrac{\hat{i} + \hat{j}}{\sqrt{2}} \times \hat{k} = \dfrac{\hat{i} - \hat{j}}{\sqrt{2}}$ , and $B_0 = E_0/c$ .
So the answer is (a) $\dfrac{E_0}{c}\dfrac{\hat{i} - \hat{j}}{\sqrt{2}}\cos\left[10^{4}\dfrac{\hat{i} + \hat{j}}{\sqrt{2}}\cdot\vec{r} - (3 \times 10^{12})t\right]$ .

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