an object is kept at distance 8cm from a converging lens of focal length 10 cm .then magnification produced by the lens is __

To find the magnification produced by a converging lens when an object is placed at a certain distance, we can use the lens formula and the magnification formula. Let's break this down step by step.

Understanding the Lens Formula

The lens formula relates the object distance (u), the image distance (v), and the focal length (f) of the lens. It is given by:

1/f = 1/v - 1/u

In this case, the focal length (f) is +10 cm (positive for converging lenses), and the object distance (u) is -8 cm (we take it as negative because, by convention, object distances are measured against the direction of the incoming light).

Calculating the Image Distance

Now, substituting the values into the lens formula:

1/10 = 1/v - 1/(-8)

This simplifies to:

1/10 = 1/v + 1/8

To combine the fractions, we find a common denominator, which is 40:

1/10 = 4/40 and 1/8 = 5/40

So, we have:

4/40 = 1/v + 5/40

Rearranging gives:

1/v = 4/40 - 5/40 = -1/40

Thus, the image distance (v) is:

v = -40 cm

This negative value indicates that the image is formed on the same side as the object, meaning it is virtual.

Finding the Magnification

The magnification (M) produced by a lens is given by the formula:

M = v/u

Substituting the values we have:

M = -40 / -8

This simplifies to:

M = 5

Interpreting the Result

A magnification of 5 means that the image is five times larger than the object. Since the magnification is positive, it indicates that the image is upright, which is typical for virtual images produced by converging lenses when the object is placed within the focal length.

Summary

In summary, when an object is placed 8 cm from a converging lens with a focal length of 10 cm, the magnification produced is 5. This means the image is five times larger than the object and is upright.