Explain image formation in a pinhole camera and calculate the height of the image given the object height and distances.

Light rays from each point of the object (the tree) travel in straight lines. These rays pass through the small pinhole. Due to the rectilinear propagation of light, rays from the top of the tree pass through the pinhole and strike the bottom of the screen, while rays from the bottom of the tree strike the top of the screen.

This process forms a real, inverted, and diminished image of the tree on the translucent screen at the back of the pinhole camera.

Object height, $H = 8 \text{ meters} = 800 \text{ cm}$
Object distance, $u = 12 \text{ meters} = 1200 \text{ cm}$
Image distance (camera length), $v = 40 \text{ cm}$

The magnification ($M$) of a pinhole camera is given by the ratio of image height ($h$) to object height ($H$), which is also equal to the ratio of image distance ($v$) to object distance ($u$):

Rearrange the formula to solve for $h$:

Substitute the values:

The size of the formed image is $\boxed{26.67 \text{ cm}}$.

If the pinhole is enlarged, more light rays from each point on the object will pass through it. Instead of a single, distinct set of rays forming a sharp image, multiple overlapping images will be formed.

This overlapping of images will cause the final image to appear blurred or less sharp. However, because more light is entering the camera, the image will also be brighter.

Moving the box (pinhole camera) closer to the tree means decreasing the object distance ($u$). The image distance ($v$, camera length) remains constant.

The magnification of a pinhole camera is inversely proportional to the object distance ($M = \frac{v}{u}$). If $u$ decreases while $v$ is constant, the magnification will increase.

Therefore, the size of the image formed on the screen will increase. The image will still be real and inverted.

Item 2

The lower fixed point ($0^\circ\text{C}$) corresponds to a liquid column length of $L_0 = 4 \text{ cm}$.
The upper fixed point ($100^\circ\text{C}$) corresponds to a liquid column length of $L_{100} = 30 \text{ cm}$.
The length of the liquid column in the bathing water is $L_\theta = 16.2 \text{ cm}$.

The temperature $\theta$ can be calculated using the formula:

Substitute the given values:

The calculated temperature of the bathing water is $46.9^\circ\text{C}$. The hint states that water meant for bathing should be between $34^\circ\text{C}$ and $38^\circ\text{C}$.
Since $46.9^\circ\text{C}$ is higher than $38^\circ\text{C}$, the water is not warm enough for bathing; it is too hot.
The water is $\boxed{\text{too hot for bathing}}$.