91Ó°ÊÓ

To determine if the image is erect or inverted, we'll think about how a pinhole camera works. In a pinhole camera, light rays from an object pass through the pinhole and create an image on the screen. Since the path of the light rays isn't changed by any lens or other optics, the image will be inverted, as the light rays from the top of the object will hit the bottom part of the screen, and those from the bottom of the object will hit the top part of the screen.

To find the magnification, we can use the ratio of the distances of the screen and the object from the pinhole: \[ M = \frac{d_\text{screen}}{d_\text{object}} \] Where \(M\) is the magnification, \(d_\text{screen}\) is the distance from the screen to the pinhole, and \(d_\text{object}\) is the distance from the object (statue) to the pinhole. Plug in the given distances: \[ M = \frac{2.8 \, \text{m}}{6.6 \, \text{m}} = \frac{14}{33} \] So the magnification is \(\frac{14}{33}\).

When we enlarge the pinhole to let more light in, the image looks blurrier because the larger pinhole diameter causes more light rays to enter from different angles, resulting in overlapping rays and a loss of sharpness in the image. This is similar to increasing the aperture size in a camera, which can cause a decrease in image sharpness if not compensated properly.

To allow more light in while still maintaining a sharp image, we should use a converging lens. A converging lens focuses the incoming light rays onto specific points on the screen, resulting in a clearer image. A diverging lens, on the other hand, would spread the light rays apart, making the image even blurrier.

To find the focal length of the lens, we can use the thin lens equation: \[ \frac{1}{f} = \frac{1}{d_\text{object}} + \frac{1}{d_\text{image}} \] Where \(f\) is the focal length, \(d_\text{object}\) is the distance from the object to the lens, and \(d_\text{image}\) is the distance from the image to the lens. Since the distance from the object to the pinhole and distance from the image to the pinhole are given as 6.6m and 2.8m, respectively, we can plug these in and solve for \(f\): \[ \frac{1}{f} = \frac{1}{6.6 \, \text{m}} + \frac{1}{2.8 \, \text{m}} \] \[ \frac{1}{f} = \frac{33 + 66}{6.6 \times 2.8} \] \[ f = \frac{6.6 \times 2.8}{99} = 0.187 \, \text{m} \] So the focal length of the lens should be approximately 0.187 meters (18.7 cm) to produce a brighter, sharper image.