A plastic cylinder of length \(3.0 \cdot 10^{1} \mathrm{~cm}\) has its ends ground to convex (from the rod outward) spherical surfaces, each having radius of curvature \(1.0 \cdot 10^{1} \mathrm{~cm}\). A small object is placed \(1.0 \cdot 10^{1} \mathrm{~cm}\) from the left end. How far will the image of the object lie from the right end, if the index of refraction of the plastic is \(1.5 ?\)

Optical physics is a branch of physics that studies the properties and behavior of light, including its interactions with matter. In optical physics, we explore phenomena such as reflection, refraction, diffraction, and dispersion. Lenses are one of the key subjects within this field, used to manipulate light in order to form images, focus beams, or correct optical imperfections.

Lenses work by bending or refracting light rays as they pass through materials of different optical densities. The way a lens bends light depends on its shape—convex lenses converge light rays, while concave lenses diverge them. Understanding how lenses affect light is crucial in fields like microscopy, photography, and even corrective eyewear.

The index of refraction, denoted as 'n', is a dimensionless number that describes how fast light travels through a material compared to the speed of light in a vacuum. It's defined as the ratio of the velocity of light in a vacuum to the velocity of light in the material:\[ n = \frac{c}{v} \]Where:

• \(c\) is the speed of light in a vacuum (approximately \(3.0 \times 10^8\) meters per second).
• \(v\) is the speed of light in the material.

A higher index means light travels slower in the material, leading to a greater bending or refraction of light as it enters or exits the material. Different materials have unique indices of refraction; for instance, our textbook problem mentions a plastic cylinder with an index of refraction of 1.5, meaning light travels slower in this plastic than in a vacuum, resulting in significant refraction at the surface.

The thin-lens formula is a relation between the focal length of a lens ('f'), the distance from the lens to the object ('d_o'), and the distance from the lens to the image ('d_i'). It is represented as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]This formula assumes that the lens is thin compared to the object and image distances, which allows us to neglect the thickness of the lens in calculations. It's a powerful tool in optical physics, used to determine where an image will form given a certain object distance, or vice versa. For example, in the problem provided, when an object is at the focal length of the spherical lens, the image forms at infinity. Conversely, any object placed at infinity will focus at the focal point of the lens on the opposite side. It's crucial in designing lenses for various applications such as cameras, telescopes, and eyeglasses. By using the thin-lens formula, we can easily calculate the necessary focal length for a desired imaging outcome.