/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 The left face of a biconvex lens... [FREE SOLUTION] | 91Ó°ÊÓ

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The left face of a biconvex lens has a radius of curvature of magnitude \(12.0 \mathrm{cm},\) and the right face has a radius of curvature of magnitude \(18.0 \mathrm{cm} .\) The index of refraction of the glass is \(1.44 .\) (a) Calculate the focal length of the lens. (b) What If? Calculate the focal length the lens has after is turned around to interchange the radii of curvature of the two faces.

Short Answer

Expert verified
The focal length of the lens in its initial orientation is 60 cm, and it is a converging lens. After the radii are interchanged, the focal length is -60 cm, and it becomes a diverging lens.

Step by step solution

01

Apply the lens maker's formula for the initial orientation

We will be using the lens maker's formula, which is given by \[1/f = (n-1) \* (1/r_1 - 1/r_2)\], where \(f\) is the focal length, \(n\) is the index of refraction, \(r_1\) is the radius of curvature for the first face (left face in our case) and \(r_2\) is the radius of curvature for the second face (right face in our case). Now, we will plug in the given values: \(n = 1.44\), \(r_1 = 12 cm\), \(r_2 = 18 cm\). When we work this out, we get \(1/f = 0.01667 cm^{-1}\). Inverting this gives the focal length \(f = 60 cm\).
02

Apply the lens maker's formula for the flipped lens

Now, we will flip the lens, interchanging the radii of curvature. Therefore, now \(r_1 = 18 cm\) and \(r_2 = 12 cm\). Plugging these values into the lens maker's formula, we get \(1/f = -0.01667 cm^{-1}\), which means the focal length is now \(f = -60 cm\).
03

Interpret the results

The focal length being positive in the first case indicates that the lens is converging while it being negative in the second case indicates that the lens is diverging.

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Key Concepts

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

Focal Length Calculation
The focal length of a lens is a measure of how strongly the lens converges or diverges light. It is the distance over which parallel rays of light are brought to a focus after passing through the lens. Calculating the focal length is crucial in the design of optical systems, such as cameras, telescopes, and eyeglasses.

In the given exercise, the lens maker's formula is used to calculate the focal length of a biconvex lens. This formula, \[1/f = (n-1) * (1/r_1 - 1/r_2)\], relates the focal length (\(f\)), the index of refraction (\(n\)), and the radii of curvature (\(r_1\) and \(r_2\)) of the lens surfaces. The solution shows how substituting the given values into the equation leads to a positive focal length, indicating a converging lens.
Biconvex Lens
A biconvex lens is symmetrical across both surfaces; it is curved outward on both sides. This shape makes the lens converging, meaning it brings light rays together to a point. Biconvex lenses are commonly used in applications where a positive focal length is required, such as magnifying glasses.

When we examine a biconvex lens, such as the one in the exercise, we assess each surface's contribution to the lens' overall power. By applying the lens maker's formula, we can determine how the curvature of each surface affects the lens' focal length. The symmetry of biconvex lenses typically simplifies calculations and optical designs.
Index of Refraction
The index of refraction, denoted as \(n\), is a fundamental property of optical materials that measures how much the material slows down light compared to its speed in a vacuum. It is this slowing down that causes light to bend or refract when entering a different medium. The index of refraction is critical in lens design because it directly affects how much the lens can bend light and thus its focal length.

In the context of our problem, the index of refraction for the lens material is \(1.44\), which is a typical value for some types of optical glass. By knowing \(n\), one can predict how lenses made of the same material will behave under various curvature conditions.
Radius of Curvature
The radius of curvature refers to the radius of the spherical cap that best fits the surface of the lens. It's a measure of how curved the lens surface is; a smaller radius means a more sharply curved surface. The radii of curvature for the two faces of a lens are instrumental in determining the lens' overall strength and focal length.

In the exercise, the two radii of curvature are given different values, 12.0 cm and 18.0 cm, for the left and right face, respectively. These values are crucial inputs in the lens maker's formula and significantly affect the outcome of the calculation. The orientation of the radii is pivotal too, as flipping the lens can alter its focusing properties, as demonstrated by the sign change in the focal length when the lens is turned around.

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Most popular questions from this chapter

An object located \(32.0 \mathrm{cm}\) in front of a lens forms an image on a screen \(8.00 \mathrm{cm}\) behind the lens. (a) Find the focal length of the lens. (b) Determine the magnification. (c) Is the lens converging or diverging?

If A spherical convex mirror has a radius of curvature with a magnitude of \(40.0 \mathrm{cm} .\) Determine the position of the virtual image and the magnification for object distances of (a) \(30.0 \mathrm{cm}\) and (b) \(60.0 \mathrm{cm} .\) (c) Are the images upright or inverted?

A concave mirror has a radius of curvature of \(60.0 \mathrm{cm}\) Calculate the image position and magnification of an object placed in front of the mirror at distances of (a) \(90.0 \mathrm{cm}\) and (b) \(20.0 \mathrm{cm} .\) (c) Draw ray diagrams to obtain the image characteristics in each case.

In some types of optical spectroscopy, such as photoluminescence and Raman spectroscopy, a laser beam exits from a pupil and is focused on a sample to stimulate electromagnetic radiation from the sample. The focusing lens usually has an antireflective coating preventing any light loss. Assume a \(100-\mathrm{mW}\) laser is located \(4.80 \mathrm{m}\) from the lens, which has a focal length of \(7.00 \mathrm{cm} .\) (a) How far from the lens should the sample be located so that an image of the laser exit pupil is formed on the surface of the sample? (b) If the diameter of the laser exit pupil is \(5.00 \mathrm{mm},\) what is the diameter of the light spot on the sample? (c) What is the light intensity at the spot?

Your friend needs glasses with diverging lenses of focal length \(-65.0 \mathrm{cm}\) for both eyes. You tell him he looks good when he doesn't squint, but he is worried about how thick the lenses will be. Assuming the radius of curvature of the first surface is \(R_{1}=50.0 \mathrm{cm}\) and the high-index plastic has a refractive index of \(1.66,\) (a) find the required radius of curvature of the second surface. (b) Assume the lens is ground from a disk \(4.00 \mathrm{cm}\) in diameter and \(0.100 \mathrm{cm}\) thick at the center. Find the thickness of the plastic at the edge of the lens, measured parallel to the axis. Suggestion: Draw a large cross-sectional diagram.

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