If the distance between the object and the pinhole of a pinhole camera is reduced by half the size of the image of the object

Analysis

To solve this problem, we need to understand the relationship between the object distance, the image distance, and the size of the image in a pinhole camera.

Concept

A pinhole camera works on the principle of the rectilinear propagation of light. The relationship between the object height ($h_o$), image height ($h_i$), object distance ($u$), and image distance ($v$) is given by the magnification formula:

Where:

• $h_i$ = size of the image
• $h_o$ = size of the object
• $v$ = distance from the pinhole to the image (fixed, as it is the length of the camera box)
• $u$ = distance from the pinhole to the object

From this formula, we can see that the size of the image ($h_i$) is inversely proportional to the object distance ($u$):

Step-by-Step Solution

1. Initial state: Let the initial object distance be $u_1$ and the initial image size be $h_{i1}$. $$h_{i1} = \frac{h_o \cdot v}{u_1}$$

2. Change in distance: The distance is reduced by half, so the new object distance $u_2$ is: $$u_2 = \frac{1}{2}u_1$$

3. New image size: Substitute $u_2$ into the formula to find the new image size $h_{i2}$: $$h_{i2} = \frac{h_o \cdot v}{u_2} = \frac{h_o \cdot v}{\frac{1}{2}u_1}$$

4. Simplify: $$h_{i2} = 2 \cdot \left( \frac{h_o \cdot v}{u_1} \right)$$ $$h_{i2} = 2 \cdot h_{i1}$$

The new image size is twice the original size.

Conclusion

When the distance between the object and the pinhole is halved, the image size becomes double the original size.

Exam Tip

Always remember the "Inverse Rule" for pinhole cameras: if you move the object closer, the image gets bigger; if you move it further away, the image gets smaller. Since the relationship is linear, if you divide the distance by 2, you multiply the size by 2.