JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 16)
A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by :
[Given permittivity of space $$ \in $$0 = 9 $$ \times $$ 10–12 SI units, Speed of light c = 3 $$ \times $$ 108 m/s]
[Given permittivity of space $$ \in $$0 = 9 $$ \times $$ 10–12 SI units, Speed of light c = 3 $$ \times $$ 108 m/s]
2 kV/m
1 kV/m
1.4 kV/m
0.7 kV/m
Explanation
Intensity of EM wave is given by
$${\rm I} = {{Power} \over {Area}} = {1 \over 2}{\varepsilon _0}E_0^2C$$
$$ = {{27 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 6}}}}$$ $$ = {1 \over 2} \times 9 \times {10^{ - 12}} \times {E^2} \times 3 \times {10^8}$$
$$E = \sqrt 2 \times {10^3}kv/m$$
$$ = 1.4$$ kv/m
$${\rm I} = {{Power} \over {Area}} = {1 \over 2}{\varepsilon _0}E_0^2C$$
$$ = {{27 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 6}}}}$$ $$ = {1 \over 2} \times 9 \times {10^{ - 12}} \times {E^2} \times 3 \times {10^8}$$
$$E = \sqrt 2 \times {10^3}kv/m$$
$$ = 1.4$$ kv/m
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