A flood light is covered with a filter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave $E_{x} = 36 \sin (1.20 \times 10^{7} z + 6 \times 10^{15} t) V / m$ . The average intensity of the beam will be

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Solution

Step1: Given data
$E_{x} = 36 \sin (1.20 \times 10^{7} z + 6 \times 10^{15} t) V / m$
Step2: Formula used
$I_{a v g} = \frac{c \times {(E_{0})}^{2} \times ε_{0}}{2} [I_{a v g} = a v e r a g e i n t e n s i t y, c = s p e e d o f l i g h t, E_{0} = a m p l i t u d e o f e l e c t r i c f i e l d, ε_{0} = e p s i l o n n a u g h t]$
Step3: Calculating average intensity
Variant electric field, $E_{x} = 36 \sin (1.20 \times 10^{7} z + 6 \times 10^{15} t) V / m$
The amplitude of the electric field, $E_{0} = 36$
Speed of light, $c = 3 \times 10^{8}$
Epsilon naught, $ε_{0} = 8.85 \times 10^{- 12} N m^{2} C^{- 2}$
Substituting the values in the formula
$I_{a v g} = \frac{c \times {(E_{0})}^{2} \times ε_{0}}{2} I_{a v g} = \frac{3 \times 10^{8} \times 8.85 \times 10^{- 12} \times {(36)}^{2}}{2} I_{a v g} = \frac{3 \times 8.85 \times 10^{- 4} \times 1296}{2} I_{a v g} = \frac{26.55 \times 1296 \times 10^{- 4}}{2} I_{a v g} = \frac{34408.8 \times 10^{- 4}}{2} I_{a v g} = 17204.4 \times 10^{- 4} I_{a v g} = 1.72 W / m^{2}$
Hence, the average intensity of the beam will be $1.72 W / m^{2}$

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Similar questions

E_{x} = 36 sin (1.20 \times 10^{7} z + 6 \times 10^{15} t) V / m

B = 12 \times 10^{- 8} sin (1.20 \times 10^{7} z - 3.6 \times 10^{15} t)

B_{0} = 12 \times 10^{- 8} sin (1.20 \times 10^{7} z - 3.60 \times 10^{15} t) T

I_{0}

^{\circ}

4 W m^{- 2}

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