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Step-by-Step Solution
Step 1: Write down the relation between intensity and electric field
For electromagnetic waves, the intensity I is related to the electric field amplitude E as:
I = \frac{1}{2} c \, \epsilon_0 \, E^2,
where c is the speed of light in vacuum, and \epsilon_0 is the permittivity of free space. From this formula, we see that
E^2 \propto I.
Step 2: Express intensity in terms of power and area
When considering a point source of power P radiating uniformly in all directions, the intensity at a distance r is
I = \frac{P}{A},
where A is the surface area of a sphere of radius r , i.e., A = 4 \pi r^2 (though the precise area cancels out in the ratio). Thus,
E^2 \propto \frac{P}{A}, \quad \text{which simplifies to} \quad E \propto \sqrt{P}.
Step 3: Relate the electric field amplitudes for two different powers
Let E be the electric field amplitude for a 100 W bulb at the same distance, and E' be the electric field amplitude for a 60 W bulb at the same distance. Since E \propto \sqrt{P} ,
\frac{E'}{E} = \sqrt{\frac{60}{100}} = \sqrt{\frac{3}{5}}.
Hence,
E' = \sqrt{\frac{3}{5}} \, E.
Step 4: Identify the value of x
The problem states that
E' = \sqrt{\frac{x}{5}} \, E.
Comparing this with the derived relation E' = \sqrt{\frac{3}{5}} \, E , we get
\sqrt{\frac{x}{5}} = \sqrt{\frac{3}{5}} \quad \Longrightarrow \quad x = 3.
Final Answer
The value of x is 3.
