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Two thin lenses of focal lengths \(f_{1}\) and \(f_{2}\) are in contact. Show that they are equivalent to a single thin lens with $$f=\frac{f_{1} f_{2}}{f_{1}+f_{2}}$$ as its focal length.

Short Answer

Expert verified
For two lenses in contact, the equivalent focal length is \( f = \frac{f_1 f_2}{f_1 + f_2} \).

Step by step solution

01

- Understand Lens Formula for Individual Lenses

The lens formula for an individual thin lens is given by \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \) where \( u \) is the object distance, \( v \) is the image distance, and \( f \) is the focal length.
02

- Combined Effect of Two Lenses in Contact

For two lenses in contact, the effective focal length \( f \) can be found using the combination formula: \( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \)
03

- Substitute Given Focal Lengths

Substitute the given focal lengths \( f_1 \) and \( f_2 \) into the formula: \( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \).
04

- Find Common Denominator

Combine the fractions on the right-hand side: \( \frac{1}{f} = \frac{f_2 + f_1}{f_1 f_2} \)
05

- Take Reciprocal of Both Sides

Taking the reciprocal of both sides to solve for \( f \), we get: \( f = \frac{f_1 f_2}{f_1 + f_2} \)

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Key Concepts

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

thin lens equation
The thin lens equation is fundamental in understanding how lenses form images. The equation is:\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]where
  • \( f \) is the focal length of the lens
  • \( u \) is the object distance from the lens
  • \( v \) is the image distance from the lens
This equation helps you determine the position of either the object or the image when you know the other two distances. Imagine a magnifying glass: you see a clear image only when the lens is at the correct distance from the object. This distance is calculated using the thin lens equation. It's crucial to remember the sign convention used in lens calculations, such as treating the focal length as positive for converging lenses and negative for diverging lenses. By mastering the thin lens equation, you can predict how a lens will behave in various situations, making it an essential part of studying optics.
focal length
The term 'focal length' is used to describe the distance between the center of a lens and the point where it converges or diverges light rays. Mathematically, it’s denoted by \( f \).The focal length is a critical feature of a lens, as it determines the lens's ability to focus light. For instance, shorter focal lengths result in more powerful lenses that can bend light rays more sharply, creating a more significant image magnification. Conversely, longer focal lengths are associated with weaker lenses that produce less magnification.When two thin lenses are placed in contact, their combined focal length \( f \) can be calculated using the formula:\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]This situation yields a single effective focal length, simplifying the system to one equivalent lens. If lenses have focal lengths of \( f_1 \) and \( f_2 \), the combined focal length becomes:\[ f = \frac{f_1 f_2}{f_1 + f_2} \]This formula shows how combining lenses can affect the overall focal length, allowing for tailored optical properties in devices like microscopes and telescopes.
optics
Optics is the branch of physics that deals with the behavior and properties of light, including its interactions with matter and the construction of instruments that use or detect it. It explores phenomena such as reflection, refraction, and diffraction.One of the most practical applications of optics is in lens design. Lenses use the principles of optics to gather light and focus it to form images. There are two main types of lenses: converging (convex) lenses and diverging (concave) lenses. Converging lenses focus incoming light rays to meet at a point, while diverging lenses spread light rays apart.In addition to forming images, lenses are essential in correcting vision, magnifying tiny objects, and capturing images in photography. Understanding how light interacts with lenses through the thin lens equation and focal length concepts is vital in fields ranging from astronomy to medical diagnostics.By combining lenses, as demonstrated in the exercise, you can manipulate the effective focal length to achieve specific magnifications and focal properties, showcasing the interplay of different optical principles to solve practical problems.

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Most popular questions from this chapter

Address each part of this question in two ways: (1) by drawing and interpreting appropriate geometrical diagrams and (2) by appealing to the lens equation and the expression for lateral magnification and demonstrating your result mathematically. If your two approaches do not agree, explain which one is correct and why the other is wrong. (a) Suppose you are using a camera and wish to have a larger image of a distant object than you are obtaining with the lens currently in use. Would you change to a lens with a longer or a shorter focal length? Explain your reasoning. (Hint: Note that the object distance is essentially fixed.) (b) Suppose you are using a slide projector and wish to obtain a larger image on the screen. You cannot achieve this by moving the screen farther from the projector because you are already using the entire length of the room. Would you change to a lens with a longer or a shorter focal length than the one you are using? Explain your reasoning. (Hint: Note that the image distance is essentially fixed.)

A converging lens with a focal length of \(+20 \mathrm{~cm}\) is located \(10 \mathrm{~cm}\) to the left of a diverging lens having a focal length of \(-15 \mathrm{~cm}\). If an object is located \(40 \mathrm{~cm}\) to the left of the converging lens, locate and describe completely the final image formed by the diverging lens.

An object is \(10.0 \mathrm{~mm}\) from the objective of a certain compound microscope. The lenses are \(300 \mathrm{~mm}\) apart and the intermediate image is \(50.0 \mathrm{~mm}\) from the eyepiece. What overall magnification is produced by the instrument?

A concave shaving mirror has a radius of curvature of \(35.0 \mathrm{~cm} .\) It is positioned so that the (upright) image of a man's face is \(2.50\) times the size of the face. How far is the mirror from the face?

Light traveling in water of refractive index \(1.33\) is incident on a plate of glass with index of refraction 1.53. At what angle of incidence is the reflected light fully polarized?

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