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The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_{1}=+12.0 \mathrm{~cm}\) and \(R_{2}=+28.0 \mathrm{~cm} .\) The index of refraction is \(1.60 .\) (a) Compute the position and size of the image of an object in the form of an arrow \(5.00 \mathrm{~mm}\) tall, perpendicular to the lens axis, \(45.0 \mathrm{~cm}\) to the left of the lens. (b) A second converging lens with the same focal length is placed \(3.15 \mathrm{~m}\) to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens \(45.0 \mathrm{~cm}\) to the right of the first.

Step by step solution

01

Calculate the Focal Length of the Lens

Use the lens-maker's formula: \(1/f = (n_{2}/n_{1} - 1) * (1/R_{1} - 1/R_{2})\), where \(n_{2}\) is the refractive index of the lens, \(n_{1}\) is the refractive index of the medium in which the lens is kept (air in this case), \(R_{1}\) and \(R_{2}\) are the radii of curvature of the lens. Replacing \(n_{2} = 1.6\), \(n_{1} = 1\), \(R_{1} = +12 \, cm\) and \(R_{2} = +28 \, cm\) in the lens-maker's formula, we compute the focal length of the lens.

02

Find the Image Position and Size for the First Lens

Use the lens equation: \(1/f = 1/v - 1/u\), where u is the object distance from the lens, v is the image distance from the lens, f is the focal length of the lens. Given \(u = -45 \, cm\), solve the equation for v to get the image position for the first lens. Then use the magnification equation \(m = -v/u\) to determine the size of the image for the first lens. The height of the image for the first lens is simply the magnification multiplied by the height of the object, which is given as \(5 \, mm\).

03

Find the Image Position and Size with Two Lenses: First Case

In this case, the second lens is \(3.15 \, m\) to the right of the first. Then consider the image produced by the first lens as the object for the second lens. And so the object distance for the second lens is simply the image distance from the first lens subtracted from the distance between the two lenses. Use the lens equation and magnification formula for the second lens as above to find the final image's position and size. The orientation of the image is determined by the sign of the magnification: if it is negative, the image is inverted; if it's positive, the image is erect.

04

Find the Image Position and Size with Two Lenses: Second Case

Here second lens is \(45 \, cm\) to the right of the first. Thus, repeat the same process as in 'Find the Image Position and Size with Two Lenses: First Case,' but now the object distance for the second lens is the image distance from the first lens subtracted from new distance between the two lenses which is \(45 \, cm\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radii of Curvature

The concept of radii of curvature is fundamental in understanding how lenses shape light. The radii of curvature refer to the radius of the spherical surfaces that make up the lens. For a lens, these are typically given as two values: one for each surface, often denoted as \(R_1\) and \(R_2\).

• In a lens, if we imagine the surfaces as sections of spheres, the radius of the sphere for each surface gives us the radii of curvature.
• In our example, \(R_1\) is given as \(+12.0\) cm and \(R_2\) as \(+28.0\) cm, both indicating that these surfaces are convex, with the center of curvature lying on the opposite side from the lens surface.
• The signs of these radii help in determining the shape: a positive value typically indicates a surface that bulges outward (convex), and a negative value indicates one that caves inward (concave).

Understanding the radii of curvature enables us to use the Lens-maker's Formula effectively, predicting how the lens will converge or diverge light.

Index of Refraction

The index of refraction, often symbolized as \(n\), describes how much a material slows down light, compared to its speed in a vacuum. This factor is crucial for lenses, as it determines how much light is bent—or refracted—as it enters and exits the lens.

• A higher index indicates a greater bending power, allowing the lens to focus light more strongly.
• In lens problems, you might encounter more than one index: \(n_1\) is the refractive index of the surrounding medium (usually air, \(n_1 = 1\)), and \(n_2\) for the lens material itself.
• In our exercise, \(n_2\) is given as \(1.60\), suggesting that the lens significantly slows down and bends light compared to air.

This value is a cornerstone in the Lens-maker's Formula, linking it with the lenses' radii of curvature to calculate their optical power.

Lens-maker's Formula

The Lens-maker's Formula is a key tool in lens optics, providing a way to calculate the focal length of a lens. Given by \(\frac{1}{f} = \left( \frac{n_2}{n_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\), it connects the lens's focal length \(f\) to its physical properties.

• \(n_2\) is the index of refraction for the lens material, and \(n_1\) is that of the surrounding medium.
• The terms \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces.
• By substituting the known values: \(n_2 = 1.6\), \(n_1 = 1\), \(R_1 = +12.0\) cm, and \(R_2 = +28.0\) cm, you can determine the focal length \(f\), which characterizes the lens's ability to focus light.

This formula is essential, as it bases the optical power of lenses not only on simple measurements but their material properties and geometrical curvature, shaping how we utilize lenses in various applications.

Image Formation

Image formation using lenses relies on precise principles of geometric optics. When light passes through a lens, it refracts at the surfaces and converges or diverges into an image.

• The position of this formed image can be predicted using the lens equation: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\), where \(u\) is the object distance and \(v\) is the image distance.
• The object's position and the focal length of the lens determine where the image will appear and whether it will be real or virtual.
• In practice, calculating \(v\) with the given \(u\) for the exercise (object at \(-45.0\) cm) and the focal length from the Lens-maker's Formula, predicts the image for the first static lens, eventually treated as an object for any following lenses in a sequence.

This method helps in visualizing and calculating the precise optical behavior and outcomes of the lenses used.

Magnification

Magnification in optics deals with how the size of the image relates to the size of the object. It's a measure of how much larger or smaller the image appears, based on the lens's properties and positioning.

• Mathematically, magnification \(m\) is given by \(m = -\frac{v}{u}\), where \(v\) is the image distance and \(u\) is the object distance.
• A magnification greater than one means the image is larger than the object, while less than one implies it's smaller.
• The negative sign in the magnification indicates image inversion. In combination with size, calculating the magnification for our lens scenario allows us to determine not just how big the image will be in comparison to the original object but also its orientation (inverted or erect).

Understanding this helps relate the abstract optical characteristics of lenses to tangible, visual changes in image scale and appearance.