The intensity of the waves from a point source at a distance [tex]$d$[/tex] from the source is [tex]$I$[/tex]. At what distance from the source is the intensity equal to [tex][tex]$2I$[/tex][/tex]?

A. [tex]\frac{d}{4}[/tex]
B. [tex]\frac{d}{\sqrt{2}}[/tex]
C. [tex]\frac{d}{8}[/tex]
D. [tex]\frac{d}{2}[/tex]



Answer :

To determine the distance from the source where the intensity of the waves is equal to [tex]\(2I\)[/tex], let's consider the relationship between intensity and distance. The intensity of waves from a point source follows the inverse square law, which states:

[tex]\[ I \propto \frac{1}{d^2} \][/tex]

where [tex]\( I \)[/tex] is the intensity and [tex]\( d \)[/tex] is the distance from the source.

Given:
- At distance [tex]\( d \)[/tex], the intensity is [tex]\( I \)[/tex].
- We need to find the new distance [tex]\( d_{\text{new}} \)[/tex] where the intensity is [tex]\( 2I \)[/tex].

Using the inverse square law, the relationship between the initial and new intensities and distances can be described as:

[tex]\[ \frac{I_{\text{new}}}{I} = \left(\frac{d}{d_{\text{new}}}\right)^2 \][/tex]

Substitute the given intensities:

[tex]\[ \frac{2I}{I} = \left(\frac{d}{d_{\text{new}}}\right)^2 \][/tex]

This simplifies to:

[tex]\[ 2 = \left(\frac{d}{d_{\text{new}}}\right)^2 \][/tex]

Taking the square root of both sides:

[tex]\[ \sqrt{2} = \frac{d}{d_{\text{new}}} \][/tex]

Solving for [tex]\( d_{\text{new}} \)[/tex]:

[tex]\[ d_{\text{new}} = \frac{d}{\sqrt{2}} \][/tex]

Therefore, the distance from the source where the intensity is equal to [tex]\( 2I \)[/tex] is:

[tex]\[ \frac{d}{\sqrt{2}} \][/tex]

Hence, the correct answer is:

[tex]\[ \boxed{\frac{d}{\sqrt{2}}} \][/tex]