Answer:
The intensity becomes one fourth when the distance is doubled.
Step-by-step explanation:
The statement is
Intensity of light [tex]l[/tex] varies inversly as the square of the distance [tex]d[/tex]
So we get
[tex]l=\frac{k}{d^2}[/tex]
Now for distance 12in let the intensity be [tex]l_{1}[/tex]
So we get
[tex]l_{1}=\frac{k}{12^2}[/tex]
[tex]l_{1}=\frac{k}{144}[/tex] ...(1)
For distance 24in, let the intensity be [tex]l_{2}[/tex]
We get
[tex]l_{2}=\frac{k}{24^2}[/tex]
[tex]l_{2}=\frac{k}{576}[/tex] ....(2)
Dividing (1) by (2) we get
[tex]\frac{l_{1}}{l_{2}}=\frac{ \frac{k}{144}}{ \frac{k}{576}}=\frac{576}{144}[/tex]
[tex]\frac{l_{1}}{l_{2}}=4[/tex]
[tex]l_{1}=4l_{2} \ or \ l_{2} = \frac{1}{4}l_{1}[/tex]
Hence the intensity becomes one fourth
