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Will the magnification produced by a simple magnifier increase, decrease, or stay the same when the object and the lens are both moved from air into water?

Short Answer

Expert verified
Answer: When the object and the lens are both moved from air into water, the magnification produced by a simple magnifier will decrease.

Step by step solution

01

Understand Magnification

Magnification (m) is defined as the ratio of the size of an image formed by a lens to the size of the object being magnified. In the case of a simple magnifier, it is given by the formula: m= 1+(D/f), where "m" is the magnification, "D" is the distance of the distinct vision (usually 25 cm), and "f" is the focal length of the lens.
02

Understand the Lens Formula and Refractive Index

The lens formula is given by: (1/f) = (n-1)(1/R1 - 1/R2), where "f" is the focal length of the lens, "n" is the refractive index of the lens material (relative to the medium in which the lens is placed), and "R1" and "R2" are the radii of curvature of the lens surfaces. When the object and lens are moved from air (n_air ≈ 1) into water (n_water ≈ 1.33), the refractive index relative to the new medium (water) will decrease, as given by the formula: n_new = n_lens/n_water, where "n_new" is the refractive index of the lens material relative to water, and "n_lens" is the refractive index of the lens material.
03

Analyze the Effect of the Change in Medium on the Focal Length

We can rewrite the lens formula for water as: (1/f_water) = (n_new-1)(1/R1 - 1/R2), As n_new < n_lens, this implies that (n_new-1) < (n_lens-1). Using the lens formula, it can be deduced that the focal length of the lens in water (f_water) will be greater than the original focal length in air (f_air).
04

Analyze the Effect of the Change in Focal Length on Magnification

From the magnification equation, we can see how the change in focal length affects magnification: m_air = 1+(D/f_air) m_water = 1+(D/f_water), As f_water > f_air, it is clear that (D/f_water) < (D/f_air). Therefore, m_water < m_air.
05

Concluding the Analysis

In conclusion, when the object and the lens are both moved from air into water, the magnification produced by a simple magnifier will decrease.

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Key Concepts

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

Understanding Magnification
Magnification is a term that comes up often in optics, especially when we're looking at simple magnifiers, such as reading glasses or jewelry loupes. In essence, magnification is the process of making an object appear larger, and in optics, it's a numerical measure of this enlargement. A key formula related to magnification for a simple magnifier is:
\[\[\begin{align*} m = 1 + \frac{D}{f},ewlineewline\end{align*}\]\]where 'm' represents the magnification, 'D' is the distance at which a healthy human eye can see an object clearly without straining, often considered to be 25 cm, and 'f' denotes the focal length of the lens. The focal length is the distance from the lens to the point where light rays converge to a focal point. With this formula, it can be observed that as the focal length ('f') decreases, the magnification ('m') increases, leading to a larger appearance of the object.In real-world applications, a simple magnifier helps people to see small details more clearly by enlarging the image when they hold an object close to their eyes. Adjusting the distance between the eye and the lens can change the focal length and therefore affect the magnification.
The Role of Refractive Index in Optics
The refractive index is a measure of how much light bends, or refracts, when it enters a material from another medium. It’s a crucial concept in optics because it determines how lenses, prisms, and other optical devices alter the path of light to create images.
The refractive index is represented by the symbol 'n' and is dimensionless. It can be understood with the formula:\[\[\begin{align*} n = \frac{c}{v},ewlineewline\end{align*}\]\]where 'c' is the speed of light in a vacuum, and 'v' is the speed of light in the material. A high refractive index indicates that light travels more slowly in the material, which means it bends more. For instance, diamond has a high refractive index, which is why it sparkles so brilliantly—the light is heavily bent and reflected inside it.When a lens is moved from air to water, the focal length changes due to the difference in refractive index of the two mediums. This influences how strongly the lens can bend light. In water, the refractive index relative to air decreases, which can result in a longer focal length, thus affecting the magnification.
Applying the Lens Formula
The lens formula is a fundamental equation in the study of optics. It mathematically relates the focal length of a lens to its refractive index and the radii of curvature of its surfaces. For any lens, the formula is written as:\[\[\begin{align*} \frac{1}{f} = (n - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right),ewlineewline\end{align*}\]\]where 'f' is the focal length, 'n' is the refractive index of the material making up the lens relative to the surrounding medium (also known as the relative refractive index), and 'R1' and 'R2' are the radii of curvature of the two surfaces of the lens. If the lens is symmetric, the radii of curvature are equal and the formula simplifies accordingly.In scenarios where the refractive index changes, such as when moving the lens from air into water, 'n' must be adjusted to reflect the new medium. If 'n' decreases, the result is an increase in focal length ('f'). This increase in focal length inversely affects the magnification, making it smaller. Hence, through the lens formula, we can predict the impact of environmental changes on the optical performance of lenses—an invaluable tool for anyone working in the field of optics or relying on precision magnification tools.

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Most popular questions from this chapter

Two distant stars are separated by an angle of 35 arcseconds. If you have a refracting telescope whose objective lens focal length is \(3.5 \mathrm{~m},\) what focal length eyepiece do you need in order to observe the stars as though they were separated by 35 arcminutes?

Galileo discovered the moons of Jupiter in the fall of \(1609 .\) He used a telescope of his own design that had an objective lens with a focal length of \(f_{\mathrm{g}}=40.0\) inches and an eyepiece lens with a focal length of \(f_{c}=2.00\) inches. Calculate the magnifying power of Galileo's telescope.

What would you expect to happen to the magnitude of the power of a lens when it is placed in water \((n=1.33) ?\) a) It would increase. d) It would depend if the lens b) It would decrease. was converging or diverging. c) It would stay the same.

An object is moved from a distance of \(30 \mathrm{~cm}\) to a distance of \(10 \mathrm{~cm}\) in front of a converging lens of focal length \(20 \mathrm{~cm}\). What happens to the image? a) Image goes from real and upright to real and inverted. b) Image goes from virtual and upright to real and inverted. c) Image goes from virtual and inverted to real and upright. d) Image goes from real and inverted to virtual and upright. e) None of the above.

You have made a simple telescope from two convex lenses. The objective lens is the one of the two lenses that is closer to the object being observed. What kind of image is produced by the eyepiece lens if the eyepiece is closer to the objective lens than the image produced by the objective lens?

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