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JEE Main 201815 AprilMorning ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Morning Shift
Mode
Online
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2018 · 15 April · Morning Shift · Online · Physics · Electromagnetic Waves

A monochromatic beam of light has a frequency v=(3)/(2 π) × 10^ 12...

A monochromatic beam of light has a frequency v=32π×1012 Hzv=\frac{3}{2 \pi} \times 10^{12} \mathrm{~Hz} and is propagating along the direction i^+j^2\frac{\hat{i}+\hat{j}}{\sqrt{2}}. It is polarized along the k^\hat{k} direction. The acceptable form for the magnetic field is:

Options

  1. AE0C(i^−j^2)cos⁡[104(i^−j^2)⋅r⃗−(3×1012)t]\frac{E_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cos \left[10^4\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cdot \vec{r}-\left(3 \times 10^{12}\right) t\right]
  2. BE0C(i^−j^2)cos⁡[104(i^+j^2)⋅r⃗−(3×1012)t]\frac{E_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cos \left[10^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \cdot \vec{r}-\left(3 \times 10^{12}\right) t\right]
  3. CE0Ck^cos⁡[104(i^+j^2)r⃗+(3×1012)t]\frac{E_0}{C} \hat{k} \cos \left[10^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \vec{r}+\left(3 \times 10^{12}\right) t\right]
  4. DE0C(i^+j^+k^)3cos⁡[104(i^+j^2)r⃗+(3×1012)t]\frac{E_0}{C} \frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}} \cos \left[10^4\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right) \vec{r}+\left(3 \times 10^{12}\right) t\right]

Correct answer

A. E0C(i^−j^2)cos⁡[104(i^−j^2)⋅r⃗−(3×1012)t]\frac{E_0}{C}\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cos \left[10^4\left(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\right) \cdot \vec{r}-\left(3 \times 10^{12}\right) t\right]

Step-by-step solution

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