JEE Main 201815 April
- Exam
- JEE Main
- Year
- 2018
- Date
- 15 April
- Subject
- Physics
- Chapter
- Electromagnetic Waves
JEE Main 2018 · 15 April · Physics · Electromagnetic Waves
A monochromatic beam of light has a frequency v = (3)/(2π) × 10 ^...
A monochromatic beam of light has a frequency and is propagating along the direction \dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}} . It is polarized along the \overset{^}{k} direction. The acceptance form the magnetic field is:
Options
- A\dfrac{{E}_{0}}{C}\left(\dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}\right)\cos \left[{10}^{4}\dfrac{\left(\overset{^}{i} + \overset{^}{j}\right)}{\sqrt{2}} . \overset{\to}{r} - \left(3 \times {10}^{12}\right)t\right]
- B\dfrac{{E}_{0}}{C}\overset{^}{k}\cos \left[{10}^{4}\dfrac{\left(\overset{^}{i} + \overset{^}{j}\right)}{\sqrt{2}} . \overset{\to}{r} + \left(3 \times {10}^{12}\right)t\right]
- C\dfrac{{E}_{0}}{C}\dfrac{\left(\overset{^}{i} - \overset{^}{j}\right)}{\sqrt{2}}\cos \left[{10}^{4}\dfrac{\left(\overset{^}{i} + \overset{^}{j}\right)}{\sqrt{2}} . \overset{\to}{r} + \left(3 \times {10}^{12}\right)t\right]
- D\dfrac{{E}_{0}}{C}\dfrac{\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right)}{\sqrt{3}}\cos \left[{10}^{4}\dfrac{\left(\overset{^}{i} + \overset{^}{j}\right)}{\sqrt{2}} . \overset{\to}{r} + \left(3 \times {10}^{12}\right)t\right]
Correct answer
C. \dfrac{{E}_{0}}{C}\dfrac{\left(\overset{^}{i} - \overset{^}{j}\right)}{\sqrt{2}}\cos \left[{10}^{4}\dfrac{\left(\overset{^}{i} + \overset{^}{j}\right)}{\sqrt{2}} . \overset{\to}{r} + \left(3 \times {10}^{12}\right)t\right]
Step-by-step solution
\overset{^}{E} \times \overset{^}{B} should have given direction of wave propagation,⇒ \overset{^}{K} \times \overset{^}{B} ∥ \dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}
\overset{^}{K} \times \left(\dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}\right) = \dfrac{\overset{^}{j} - \left(- \overset{^}{i}\right)}{\sqrt{2}} = \dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}} ∥ \dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}.
Wave propagation vector \overset{^}{K} should be along \dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}.
\overset{^}{K} \times \left(\dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}\right) = \dfrac{\overset{^}{j} - \left(- \overset{^}{i}\right)}{\sqrt{2}} = \dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}} ∥ \dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}.
Wave propagation vector \overset{^}{K} should be along \dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}.