A point source is fixed 1.0 m away from a large screen. Call the line normal to the screen surface and passing through the center of the point source the z axis. When a sheet of cardboard in which a square hole 0.020 m on a side has been cut is placed between the point source and the screen, 0.50 m from the point source with the hole centered on the z axis, a bright square shows up on the screen. If, instead, a second sheet of cardboard with a similar square hole is placed between the point source and screen, 0.25 m from the point source with the hole centered on the z axis, the bright square it casts on the screen is identical to the bright square from the first sheet. What is the length of the side of the hole in this sheet?

To find the length of the side of the hole in the second sheet, we can use the similarities in the geometry created by both the sheets of cardboard. Because the point source of light is fixed 1.0 m from a large screen, as light travels from the source to the screen through the holes in the sheets of cardboard, it creates larger squares on the screen.

Each square on the screen can be regarded as an image of the corresponding square hole in the cardboard, magnified as the light rays diverge from the point source. The fact that identical bright squares are cast on the screen by both sheets indicates that this magnification is the same for both.

Let's denote the side length of the hole in the first sheet as \( L_1 = 0.020 \) m and the side length of the hole in the second sheet as \( L_2 \).

Both squares are 1.0 m away from the point source, and the distances from the source to the cardboards are given as \( D_1 = 0.50 \) m for the first sheet and \( D_2 = 0.25 \) m for the second sheet.

Since the size of the bright squares on the screen is the same from both cardboards, we can write the following proportion based on similar triangles:

\[ \frac{L_1}{D_1 + 1.0} = \frac{L_2}{D_2 + 1.0} \]

This can be simplified to:

\[ \frac{0.020 \text{ m}}{0.50 \text{ m} + 1.0 \text{ m}} = \frac{L_2}{0.25 \text{ m} + 1.0 \text{ m}} \]

Solving this equation for \( L_2 \), we get:

\[ L_2 = \frac{0.020 \text{ m} \times (0.25 \text{ m} + 1.0 \text{ m})}{0.50 \text{ m} + 1.0 \text{ m}} \]

Putting in the numbers:

\[ L_2 = \frac{0.020 \text{ m} \times 1.25 \text{ m}}{1.50 \text{ m}} \]

\[ L_2 = 0.020 \text{ m} \times \frac{5}{6} \]

\[ L_2 = \frac{0.020 \text{ m} \times 5}{6} \]

\[ L_2 = \frac{100}{6} \times 10^{-3} \text{ m} \]

\[ L_2 = \frac{50}{3} \times 10^{-3} \text{ m} \]

\[ L_2 = 16.67 \times 10^{-3} \text{ m} \]

So the length of the side of the hole in the second sheet of cardboard is 0.01667 m.

Extra: The situation described is a basic illustration of geometric optics. In geometric optics, when light passes through an aperture (like the square hole in the cardboard), it can be thought of as radiating out from the aperture in every direction. This spread of light creates an image of the aperture at some distance away.

The relationship between the distance of an object (in this case, the square hole) from a point source and the size of the shadow or image it creates can usually be simplified into similar triangles, allowing for quick and easy calculations.

Furthermore, if the holes were much smaller or the point source were a coherent source (like a laser), we would have to consider diffraction effects, which can dramatically change the shape and intensity of light on the screen, leading to patterns and bright spots known as diffraction patterns. However, in the scenario described, since the holes are relatively large and there's no mention of the light being coherent, we are justified in ignoring diffraction and simply using the rules of geometric optics