Answer :
Answer:
Intensity at 45 cm will be 0.2283 volt
Explanation:
We have given that intensity of signal measured by the detector is 0.74 volt at a distance of 25 cm
We know that intensity of signal is given by [tex]I=\frac{s}{r^2}[/tex] , here s is strength of light and r is distance
So [tex]0.74=\frac{s}{{25}^2}[/tex]
[tex]s=462.5[/tex]
In second case
Intensity is given by [tex]I=\frac{s}{r^2}[/tex]
So [tex]I=\frac{462.5}{45^2}=0.2283volt[/tex]