Question

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 \(1.5 ?\)

Short answer

Answer: The image is formed at a distance of \(1.5 \cdot 10^{1} \mathrm{~cm}\) from the right end of the cylinder.

Step-by-step solution

Apply the lens maker's equation to the left end of the cylinder

The lens maker's equation is given by: \( \frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \) Where: - \(f\) is the focal length of the lens - \(n\) is the index of refraction - \(R_1, R_2\) are the radii of curvature of the two surfaces Since both spherical surfaces are convex (outward), \(R_1\) will be positive and \(R_2\) will be negative: \( R_1 = 1.0 \cdot 10^{1} \mathrm{~cm} \) \( R_2 = -1.0 \cdot 10^{1} \mathrm{~cm} \) We are given that the index of refraction of the plastic is \(1.5\). Now, we can calculate the focal length of the left end: \( \frac{1}{f} = (1.5 - 1) \left( \frac{1}{1.0 \cdot 10^{1}} - \frac{1}{-1.0 \cdot 10^{1}}\right)\) \( \frac{1}{f} = 0.5 \left( \frac{1}{1.0 \cdot 10^{1}} + \frac{1}{1.0 \cdot 10^{1}}\right)\) \( \frac{1}{f} = 0.5 \cdot \frac{2}{1.0 \cdot 10^{1}}\) \( f = \frac{1}{0.5 \cdot \frac{2}{1.0 \cdot 10^{1}}}\) \( f = 1.0 \cdot 10^{1} \mathrm{~cm}\) The focal length of the left end of the cylinder is \(1.0 \cdot 10^{1} \mathrm{~cm}\).

Apply the lens maker's equation to the right end of the cylinder

Since the focal length and the radii of curvature are the same for the right end of the cylinder, we don't need to repeat the calculation. The focal length of the right end of the cylinder is the same as the left end, \(f = 1.0 \cdot 10^{1} \mathrm{~cm}\).

Apply the thin-lens formula to find the position of the image on the left end

The thin-lens formula is given by: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) Where: - \(d_o\) is the distance of the object from the lens - \(d_i\) is the distance of the image from the lens We are given that the object is placed at a distance of \(1.0 \cdot 10^{1} \mathrm{~cm}\) from the left end. Now, we can calculate the position of the image on the left end: \( \frac{1}{1.0 \cdot 10^{1}} = \frac{1}{1.0 \cdot 10^{1}} + \frac{1}{d_i}\) \( \frac{1}{d_i} = 0 \) \( d_i = \infty\) The image on the left side of the cylinder is formed at infinity.

Calculate the position of the corresponding object on the right side of the cylinder

Since the image formed on the left end is at infinity, the object for the right end will be at a distance of \(3.0 \cdot 10^{1} \mathrm{~cm}\) (the length of the cylinder).

Apply the thin-lens formula to find the position of the image on the right end

Using the thin-lens formula with \(d_o = 3.0 \cdot 10^{1} \mathrm{~cm}\) for the right end: \( \frac{1}{1.0 \cdot 10^{1}} = \frac{1}{3.0 \cdot 10^{1}} + \frac{1}{d_i}\) \( \frac{1}{d_i} = \frac{1}{1.0 \cdot 10^{1}} - \frac{1}{3.0 \cdot 10^{1}}\) \( \frac{1}{d_i} = \frac{2}{3.0 \cdot 10^{1}}\) \( d_i = \frac{3.0 \cdot 10^{1}}{2}\) \( d_i = 1.5 \cdot 10^{1} \mathrm{~cm}\) The image of the object lies at a distance of \(1.5 \cdot 10^{1} \mathrm{~cm}\) from the right end of the cylinder.

Key concepts

Optical Physics

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.

Index of Refraction

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.

Thin-Lens Formula

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.