If the intensity of an X-ray becomes I₀/3 form I₀ after travelling 3.5 cm inside a target, then its intensity after travelling next 7 cm will be how much?

Dear Student,

Please find below the solution to the asked query:

Intensity decreases with distance as in the following relation,
$I \alpha \frac{1}{r^2}$
So, Given that at a distance 3.5 cm, intensity is $\frac{I_0}{3}$, where I₀ is initial intensity.
Therefore,
$$\begin{gathered} I \alpha \frac{1}{r^2} \Rightarrow I = \frac{K}{r^2} \Rightarrow \frac{I_0}{3} = \frac{K I_0}{{\left(3.5\right)}^2} \\ \Rightarrow K = \frac{{\left(3.5\right)}^2}{3} \end{gathered}$$
So, intensity at a distance of 7 cm is,
$$\begin{gathered} I\text{'} = K\frac{I_0}{{\left(7\right)}^2}=\left(\frac{{\left(3.5\right)}^2}{3}\right)\left(\frac{I_0}{{\left(7\right)}^2}\right)=\left(\frac{{\left(3.5\right)}^2}{3}\right)\left(\frac{I_0}{4{\left(3.5\right)}^2}\right) \\ \Rightarrow I\text{'} = \frac{I_0}{12} \end{gathered}$$


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