To find the image position, size, and nature for a converging lens, we use the lens formula 1/f = 1/v − 1/u and the magnification formula m = v/u = hi/ho. Given f = 10 cm and u = −25 cm (object distance is negative), solving the lens formula gives v ≈ 16.67 cm (positive, so the image is real and on the opposite side of the lens). The magnification m ≈ −0.67, indicating the image is inverted and smaller (about 3.35 cm in length).


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Introduction to Image Formation by Converging Lenses

Basic Principles of Lens Optics: Converging lenses, or convex lenses, are capable of bending light rays to meet at a point. The nature of the image formed by these lenses depends on the object’s distance from the lens, the lens’s focal length, and the lens’s optical properties.

Given Scenario

Object and Lens Specifications: In this scenario, an object 5 cm in length is placed 25 cm away from a converging lens with a focal length of 10 cm. The objective is to determine the position, size, and nature of the image formed by the lens.

Calculating Image Position

Using the Lens Formula: The lens formula 1/f = 1/v − 1/u is used to find the image distance v. Here, f is the focal length and u is the object distance. With f = 10 cm and
u = −25 cm (object distance is taken as negative), solving the formula gives v ≈ 16.67 cm.

Determining Image Size

Applying the Magnification Formula: The magnification formula m v/u = hi/ho relates the image height hi to the object height ho. With m ≈ −0.67 (from v ≈ 16.67 cm and u = −25 cm), the image height is found to be approximately 3.35 cm, smaller than the object.

Nature of the Image

Real and Inverted: The positive value of v indicates that the image is real and formed on the opposite side of the lens. The negative magnification indicates that the image is inverted relative to the object.

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Summary of Image Characteristics: Therefore, the image formed by the converging lens is located approximately 16.67 cm from the lens, is about 3.35 cm in length, and is real and inverted. This example illustrates the fundamental principles of image formation in convex lenses and their practical applications in optics.

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Questions of 10th Science Chapter 9 in Detail

Define the principal focus of a concave mirror.
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Define 1 dioptre of power of a lens.
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Find the power of a concave lens of focal length 2 m.
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An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. What is the position, size and the nature of the image formed.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens?
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
The magnification produced by a plane mirror is +1. What does this mean?
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained?
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?