Question

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Answer 1

Answer:

The intensity becomes one fourth when the distance is doubled.

Step-by-step explanation:

The statement is

Intensity of light $l$ varies inversly as the square of the distance $d$

So we get

$l=\frac{k}{d^2}$

Now for distance 12in let the intensity be $l_{1}$

So we get

$l_{1}=\frac{k}{12^2}$

$l_{1}=\frac{k}{144}$    ...(1)

For distance 24in, let the intensity be $l_{2}$

We get

$l_{2}=\frac{k}{24^2}$

$l_{2}=\frac{k}{576}$       ....(2)

Dividing (1) by (2) we get

$\frac{l_{1}}{l_{2}}=\frac{ \frac{k}{144}}{ \frac{k}{576}}=\frac{576}{144}$

$\frac{l_{1}}{l_{2}}=4$

$l_{1}=4l_{2} \ or \ l_{2} = \frac{1}{4}l_{1}$

Hence the intensity becomes one fourth