A converging lens with a focal length of 90.0 cm forms an image of a 3.20-cm -tall real object

A converging lens with a focal length of 90.0 cm forms an...

A converging lens with a focal length of $90.0 \mathrm{~cm}$ forms an image of a $3.20-\mathrm{cm}$ -tall real object that is to the left of the lens. The image is $4.50 \mathrm{~cm}$ tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

Key Concepts

Converging Lenses

Converging lenses, also known as convex lenses, have a positive focal length and are designed to bend light rays inward. They are used to form real images when the object is placed at a distance from the lens greater than the focal length, and they can also produce virtual images when the object is within the focal length.

Thin Lens Equation

This equation, commonly written as 1/f = 1/d? + 1/d?, links the focal length of a lens with the distances from the lens to the object (d?) and the image (d?). It is fundamental in geometrical optics for predicting where an image will form given a certain object placement.

Magnification

Magnification describes the ratio of the image height to the object height and is also given by -d?/d?. This concept is crucial for determining not only the size of the image relative to the object but also its orientation, with a negative sign indicating the image is inverted relative to the object.

Real vs Virtual Images

Real images occur when light rays actually converge at a point after passing through a lens, allowing the image to be projected onto a screen. In contrast, virtual images are formed when the light rays diverge, giving the appearance of an image at a location where the light does not actually converge.

Transcript

00:02 Hi, in the given problem, the focal length of the convex lens is given as 90 .0 centimeter and as a convex lens is a converging lens, so its focal length will be taken as positive.

00:17 Now, the height of object kept in front of it is given as 3 .20 centimeter and as the object is always erect, so it is taken as positive.

00:30 While the inverted image is given as having a height of 4 .50 centimeter and as it is inverted, so its height is taken as negative with the help of sign convention.

00:45 Now we have to find the object distance and the image distance in relation to the position of this convex lens.

00:56 So first of all, using things.

01:00 In lens equation, which is a relation among the object distance, image distance, and focal length of a lens.

01:11 And it says 1 by pi minus 1 by po is equal to 1 by f.

01:19 So now we multiply both the sides by object distance po.

01:24 So it becomes po by pi minus po by po is equal to po by f.

01:35 So it may be written as po by pi minus 1 is equal to po by f or finally we can give it as p o by pi is equal to 1 plus po by f or making lcm it becomes f plus po by f.

01:59 Now using the expression for linear magnification of a lens.

02:09 So this linear magnification produced by a lens is given as the ratio of height of the image to the height of the object and it is also proved to be equal to the ratio of image distance means pi, to the object distance means po.

02:29 So using this, it can be written as f by f plus po...

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