7 fill in the blank 2 points if distance is halved, how much is the beam intensity affected? would image…

Explanation:

Step1: Recall the inverse - square law

The intensity $I$ of a beam is inversely proportional to the square of the distance $d$ from the source, i.e., $I=\frac{k}{d^{2}}$ where $k$ is a constant.

Step2: Change the distance and find new intensity

Let the initial distance be $d_1$ and intensity be $I_1=\frac{k}{d_1^{2}}$. If the distance is halved, so $d_2 = \frac{d_1}{2}$. Then the new intensity $I_2=\frac{k}{d_2^{2}}=\frac{k}{(\frac{d_1}{2})^{2}}=\frac{k}{\frac{d_1^{2}}{4}} = 4\times\frac{k}{d_1^{2}}$. So $I_2 = 4I_1$.

Step3: Determine effect on image density

Since beam intensity is related to image density (higher intensity means more photons hitting the imaging surface per unit area), and the intensity has increased, the image density will increase.

Answer:

The beam intensity is quadrupled. Image density increases.