The electric field intensity produced by the radiations coming from 100W bulbs at a 3m distance is E. The electric field intensity produced by the radiations coming from 50W bulb at the same distance is

Options

A. `E/2`
B. `2E,`
C. `E/(sqrt(2))`
D. `sqrt(2)E`

Answer

A

Explanation

IF P is the power emitted by bulb then electric field intensity on a surface of area A due to incident radiation is
` E=(P)/(A )`
if A is constant (( area of a surface ) , then
` E prop P`
` therefore (E_1)/(E_2) = (P_1)/(P_2)`
where ` E_1 =E`
`P_1 = 100 W `
`P_2 =50 W`
`implies E_2 =(P_2 E_1)/(P_1) =( 50)/(100 ) xx E =(E )/(2)`