/*! 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 46 The focal points of a thin diver... [FREE SOLUTION] | 91Ó°ÊÓ

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The focal points of a thin diverging lens are \(25.0 \mathrm{~cm}\) from the center of the lens. An object is placed to the left of the lens, and the lens forms an image of the object that is \(18.0 \mathrm{~cm}\) from the lens. (a) Is the image to the left or right of the lens? (b) How far is the object from the center of the lens? (c) Is the height of the image less than, greater than, or the same as the height of the object?

Short Answer

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(a) The image is to the left of the lens. (b) The object is 45.0 cm from the center of the lens. (c) The height of the image is less than the height of the object.

Step by step solution

01

Determine the type of image

By property of the thin diverging lens, the image is virtual and upright. Therefore it is located on the same side as the object i.e., on the left side of the lens.
02

Calculate the object distance

We can use the lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\), where \(f\) is the focal length, \(v\) is the image distance (negative for virtual images on same side as object), and \(u\) is the object distance. Substituting \(f = -25.0 \, cm\) and \(v = -18.0 \, cm\), we can solve for \(u\). Rearranging gives us \(\frac{1}{u} = \frac{1}{f} + \frac{1}{v}\). Calculating gives us \(u = 45.0 \, cm\). Therefore, the object is located 45.0 cm from the lens.
03

Compare the height of image and object

Diverging lens form image that are always smaller than original object. Hence the height of the image is less than the height of the object.

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Key Concepts

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

Lens Formula
The lens formula is a fundamental aspect of optics that connects the distances of the object, the image, and the focal length of a lens. It is represented as \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length of the lens, \( v \) is the image distance from the lens, and \( u \) is the object distance from the lens.

For diverging lenses, the focal length is taken as negative because diverging lenses spread out light rays, implying that their focus is virtual and on the same side as the object. Similarly, for virtual images formed by such lenses, the image distance \( v \) is also negative.

To apply the lens formula, precisely understanding the signs in the context of the lens being used is essential. By rearranging and solving the formula, you can find an unknown value when the other two are provided.
Optics
Optics is the branch of physics that deals with the study of light and its interactions with matter. In the context of lenses, optics examines how light rays are refracted when they pass through different mediums and how they form images.

Refraction and Lenses

The bending of light as it passes from one medium to another is known as refraction, and it’s the principle behind lenses focusing light. Diverging lenses, which have a thicker edge than the center, cause light rays to spread out, effectively extending their path.

Real and Virtual Images

In optics, images formed by lenses can be real or virtual. Real images are formed when light rays actually converge, while virtual images, such as those formed by a thin diverging lens, appear where the light rays, if extended backward, appear to diverge from.
Virtual Images
Virtual images are formed when the rays diverge after striking a lens or mirror and cannot be projected onto a screen. For diverging lenses, the image can always be found on the same side as the object.

In our exercise, because the lens is diverging and the image forms on the same side as the object, the image distance is negative, signifying that it is virtual. Virtual images are upright, and although they can be viewed through the lens, they are, simply, an optical illusion of the actual object not actually present in the space where they appear to be.
Focal Length
In the world of optics, focal length is the distance from the center of a lens to its focal point, the location at which parallel rays of light converge (for converging lenses) or appear to diverge from (for diverging lenses).

The focal length determines the lens's power and is instrumental in calculating the object and image distances using the lens formula. A longer focal length implies a weaker lens (more subtle bending of light) and vice versa. In the case of the diverging lens from our exercise, the focal length is negative, indicating that it's on the same side of the lens as the object and exemplifying its nature to spread out light rather than bringing it to a focus.

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Most popular questions from this chapter

A camera with a 90-mm-focal-length lens is focused on an object \(1.30 \mathrm{~m}\) from the lens. To refocus on an object \(6.50 \mathrm{~m}\) from the lens, by how much must the distance between the lens and the sensor be changed? To refocus on the more distant object, is the lens moved toward or away from the sensor?

In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of \(1.40,\) and all the bending occurs at the cornea, whose vertex is \(2.60 \mathrm{~cm}\) from the retina. What should be the radius of curvature of the cornea such that the image of an object \(40.0 \mathrm{~cm}\) from the cornea's vertex is focused on the retina?

The left end of a long glass rod \(8.00 \mathrm{~cm}\) in diameter, with an index of refraction of 1.60 , is ground and polished to a convex hemispherical surface with a radius of \(4.00 \mathrm{~cm}\). An object in the form of an arrow \(1.50 \mathrm{~mm}\) tall, at right angles to the axis of the rod, is located on the axis \(24.0 \mathrm{~cm}\) to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

A small tropical fish is at the center of a water-filled, spherical fish bowl \(28.0 \mathrm{~cm}\) in diameter. (a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?

A thin lens with a focal length of \(6.00 \mathrm{~cm}\) is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, \(25.0 \mathrm{~cm}\) from the eye, and that the lens is very close to the eye.

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