Question

Intensity of radiation received by an absorber
is 100 units when the distance between source
& absorber is 'd' units. If the distance is
doubled then intensity received will be​

Answer 1

Answer:

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Answer 2

Given:

Intensity of radiation at distance 'd' $I_{0}=100 units$

Initial distance $r=d$

Final distance $r'=2d$

To Find:

Intensity of radiation at distance "2d" is $I'=?$

Formula:

$I=\frac{k}{r^{2} }$

Solution:

Step 1 of 2

Intensity of radiation emitting from any source obeys inverse square law.

Inverse square law states that the intensity is inversely proportional to the square of the distance of separation between source and absorber.

If r is the distance between source and absorber and I is the intensity, then

$I\propto\frac{1}{r^{2} }$

The proportionality symbol is replaced to equal to sign by inserting a proportionality constant say, k.

$I=\frac{k}{r^{2} }$

Step 2 of 2

If at distance "d" the intensity is 100, means

$I_{0}=\frac{1}{d^{2} } =100$

Then, if the distance is doubled, the intensity will be found as

$I'=\frac{1}{(2d)^{2} }=\frac{1}{4d^{2} }$

Substitute 100 in the place of $1/d^{2}$ in the above equation, we get

$I'=\frac{100}{4}=25 units$

So, the intensity of radiation when the distance is doubled is 25 units.

Final Answer:

If the distance is doubled then the intensity received will be 25 units.