A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t = t1 it is found that the electric field is zero at a spatial point z1. The next zero that occurs in its neighbourhood is at z2. The frequency of the electromagnetic wave is :

Question

A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t = t1 it is found that the electric field is zero at a spatial point z1. The next zero that occurs in its neighbourhood is at z2. The frequency of the electromagnetic wave is : (A) (3 × 108/|z2 - z1|) (B) (1.5 × 108/|z2 - z1|) (C) (6 × 108/|z2 - z1|) (D) (1/t1 + (|z2 - z1|)3 × 108)

Tardigrade Admin · Accepted Answer

Since c=f λ ⇒ f=(c/λ). Here, λ=2|z2-z1| and c=3 × 108 . Therefore, f=(3 × 108/2|z2-z1|)=(1.5 × 108/|z2-z1|)

Q. A plane polarized monochromatic $EM$ wave is traveling in vacuum along $z$ direction such that at $t = t_{1}$ it is found that the electric field is zero at a spatial point $z_{1}$ . The next zero that occurs in its neighbourhood is at $z_{2}$ . The frequency of the electromagnetic wave is :

$\frac{3 \times 1 0 ^{8}}{∣ z _{2} - z _{1} ∣}$

$\frac{1.5 \times 1 0 ^{8}}{∣ z _{2} - z _{1} ∣}$

$\frac{6 \times 1 0 ^{8}}{∣ z _{2} - z _{1} ∣}$

$\frac{1}{t _{1} + \frac{∣ z _{2} - z _{1} ∣}{3 \times 1 0 ^{8}}}$

Since $c = f λ \Rightarrow f = \frac{c}{λ}$ .
Here, $λ = 2 ∣ z_{2} - z_{1} ∣$ and $c = 3 \times 1 0^{8} .$ Therefore,
$f = \frac{3 \times 1 0 ^{8}}{2 ∣ z _{2} - z _{1} ∣} = \frac{1.5 \times 1 0 ^{8}}{∣ z _{2} - z _{1} ∣}$

Q. A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t=t1​ it is found that the electric field is zero at a spatial point z1​. The next zero that occurs in its neighbourhood is at z2​. The frequency of the electromagnetic wave is :

∣z2​−z1​∣3×108​

∣z2​−z1​∣1.5×108​

∣z2​−z1​∣6×108​

t1​+3×108∣z2​−z1​∣​1​