JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 1)

A monochromatic beam of light has a frequency $$v = {3 \over {2\pi }} \times {10^{12}}Hz$$ and is propagating along the direction $${{\widehat i + \widehat j} \over {\sqrt 2 }}.$$
It is polarized along the $$\widehat k$$ direction. The acceptable form for the magnetic field is :
JEE Main 2018 (Online) 15th April Morning Slot Physics - Electromagnetic Waves Question 140 English Option 1
JEE Main 2018 (Online) 15th April Morning Slot Physics - Electromagnetic Waves Question 140 English Option 2
JEE Main 2018 (Online) 15th April Morning Slot Physics - Electromagnetic Waves Question 140 English Option 3
JEE Main 2018 (Online) 15th April Morning Slot Physics - Electromagnetic Waves Question 140 English Option 4

Explanation

Given : Frequency $$f = {3 \over {2\pi }} \times {10^{12}}$$ Hz; direction $$ = {{\widehat i + \widehat j} \over {\sqrt 2 }}$$.

The beam is polarised along $${\widehat k}$$ direction.

Direction of magnetic field $$\overrightarrow B = \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }} \times \widehat k = {{ - \widehat j} \over {\sqrt 2 }} + {{\widehat i} \over {\sqrt 2 }}} \right)$$

Therefore, direction of $$\overrightarrow B = \left( {{{\widehat i - \widehat j} \over {\sqrt 2 }}} \right)$$

Now from wave equation, we have

$$B = {B_0}\cos (kr - \omega t)$$

$$\omega = 2\pi f = 2\pi \times {3 \over {2\pi }} \times {10^{12}}Hz = 3 \times {10^{12}}Hz$$

Now, $$k = {{\widehat i + \widehat j} \over {\sqrt 2 }}$$;

$${B_0} = {{{E_0}} \over c}\left( {{{\widehat i - \widehat j} \over {\sqrt 2 }}} \right)$$

Therefore,

$$B = {{{E_0}} \over c}\left( {{{\widehat i - \widehat j} \over {\sqrt 2 }}} \right)\cos \left[ {{{10}^4}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)\,.\,r - (3 \times {{10}^{12}})t} \right]$$

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