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Physics Topics are often described using mathematical equations, making them precise and quantifiable.

## Sign convention for Spherical Lenses and Lens Formula

These days New Cartesian Sign Convention is used for measuring the various distances in the ray diagrams of spherical lenses (convex lenses and concave lenses). According to the New Cartesian Sign Convention:

- All the distances are measured from the optical centre of the lens.
- The distances measured in the same direction as that of incident light are taken as positive.
- The distances measured against the direction of incident light are taken as negative.
- The distances measured upward and perpendicular to the principal axis are taken as positive.
- The distances measured downward and perpendicular to the principal axis are taken as negative.

The New Cartesian Sign Convention for lenses is shown in Figure. The object is always placed on the left side of the lens (as shown in Figure), so that the direction of incident light is from left to right. All the distances measured from the optical centre (C) of the lens to the right side will be considered positive (because they will be in the same direction as the incident light). On the other hand, all the distances measured from the optical centre (C) of the lens to the left side are considered negative (because they are measured against the direction of incident light).

On the basis of New Cartesian Sign Convention, the focal length of a convex lens is considered positive (and written with a plus sign). On the other hand, the focal length of a concave lens is considered negative (and written with a minus sign). Please note that the sign convention for spherical lenses is very similar to the sign convention for spherical mirrors which we have already studied.

### What is Lens Formula

A formula which gives the relationship between image distance (v), object distance (u), and focal length if) of a lens is known as the lens formula. The lens formula can be written as :

\(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)

where v = image distance

u = object distance and

f = focal length

This lens formula applies to both types of spherical lenses : convex lenses as well as concave lenses. Please note that the lens formula differs from the mirror formula only in the sign between \(\frac{1}{v}\) and \(\frac{1}{u}\). The mirror formula is \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\) whereas the lens formula is \(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\). It is clear that the mirror formula has a plus sign (+) between and whereas the lens formula has a minus sign (-) between and The values of v’ and u should be substituted in the lens formula with their proper signs.

### Magnification Produced by Lenses

The size of image formed by a lens depends on the position of the object from the lens. For example, the image formed by a convex lens can be smaller than the object, equal to the object or bigger than the object. The size of the image relative to the object is given by the linear magnification. The linear magnification is the ratio of the height of the image to the height of the object. That is:

Magnification \(=\frac{\text { height of image }}{\text { height of object }}\)

or m = \(\frac{h_2}{h_1}\)

where m = magnification

h_{2} = height of image and h_{1} = height of object

We will now write another formula for the magnification produced by a lens in terms of the image distance and object distance.

The linear magnification produced by a lens is equal to the ratio of image distance to the object

distance. That is :

Magnification \(=\frac{\text { Image distance }}{\text { Object distance }}\)

or m = \(\frac{v}{u}\)

where m = magnification

v = image distance

and u = object distance

It should be noted that this second magnification formula for lenses differs only in sign from the magnification formula for mirrors. The magnification formula for the mirrors is, m = \(\frac{v}{u}\) whereas that for lenses is, m = \(\frac{v}{u}\). It is clear that the magnification formula for mirrors has a minus (-) sign but the magnification formula for lenses has no minus sign.

If the magnification m has a positive value, the image is virtual and erect. And if the magnification m has a negative value, the image will be real and inverted. Since a convex lens can form virtual images as well as real images, therefore, the magnification produced by a convex lens can be either positive or negative. A concave lens, however, forms only virtual images, so the magnification produced by a concave lens is always positive. A convex lens can form images which are smaller than the object, equal to the object or bigger than the object, therefore, the magnification (m) produced by a convex lens can be less than 1, equal to 1 or more than 1, On the other hand, a concave lens forms images which are always smaller than the object, so the magnification (m) produced by a concave lens is always less than 1.

### Numerical Problems Based On Convex Lenses

We will now solve some numerical problems based on convex lenses by using the lens formula and the magnification formulae. Here are some examples.

**Example Problem 1.**

A convex lens of focal length 10 cm is placed at a distance of 12 cm from a wall. How far from the lens should an object be placed so as to form its real image on the wall ?

**Solution.**

Here, the real image is formed on the wall which is at a distance of 12 cm from the convex lens. This means that the distance of image from the convex lens or image distance will be 12 cm. Since a real image is formed on the right side of the lens, so this image distance will be positive.

Now, Image distance, v = + 12 cm (A real image)

Object distance, u = ? (To be calculated)

Focal length, f = + 10 cm (It is a convex lens)

Putting these values in the lens formula :

\(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)

we get : \(\frac{1}{12}\) – \(\frac{1}{u}\) = \(\frac{1}{10}\)

\(\frac{1}{12}\) – \(\frac{1}{10}\) = \(\frac{1}{u}\)

\(\frac{5-6}{60}\) = \(\frac{1}{u}\)

– \(\frac{1}{60}\) = \(\frac{1}{u}\)

So, Object distance, u = – 60 cm

Thus, the object should be placed at a distance of 60 cm in front of the convex lens. The minus sign shows that the object is on the left side of the lens.

**Example Problem 2.**

If an object of 7 cm height is placed at a distance of 12 cm from a convex lens of focal length 8 cm, find the position, nature and height of the image.

**Solution.**

First of all we will find out the position of the image. By the position of image we mean the distance of image from the lens.

Here, Object distance, u = -12 cm (It is to the left of lens)

Image distance, v = ? (To be calculated)

Focal length, f = + 8 cm (It is a convex lens)

Putting these values in the lens formula :

\(\frac{1}{v}\) – \(\frac{1}{u}\) = \(\frac{1}{f}\)

we get : \(\frac{1}{v}\) – \(\frac{1}{-12}\) = \(\frac{1}{8}\)

or \(\frac{1}{v}\) + \(\frac{1}{12}\) = \(\frac{1}{8}\)

\(\frac{1}{v}\) = \(\frac{1}{8}\) – \(\frac{1}{12}\)

\(\frac{1}{v}\) = \(\frac{3-2}{24}\)

\(\frac{1}{v}\) = \(\frac{1}{24}\)

So, Image distance, v = + 24 cm

Thus, the image is formed at a distance of 24 cm from the convex lens. The plus sign for image distance shows that the image is formed on the right side of the convex lens. Only a real and inverted image is formed on the right side of a convex lens, so the image formed is real and inverted.

Let us calculate the magnification now. We know that for a lens :

Magnification, m = \(\frac{v}{u}\)

Image distance, v = 24 cm

Object distance, u = -12 cm

So, m = \(\frac{24}{-12}\)

or m = – 2

Since the value of magnification is more than 1 (it is 2), so the image is larger than the object or magnified. The minus sign for magnification shows that the image is formed below the principal axis. Hence the image is real and inverted. Let us calculate the size of the image by using the formula :

m = \(\frac{h_2}{h_1}\)

Here, Magnification, m = – 2 (Found above)

Height of object, h_{1} = + 7 cm (Measured upwards)

Height of image, h_{2} = ? (To be calculated)

Now, putting these values in the above formula, we get:

-2 = \(\frac{h_2}{7}\)

or h_{2} = – 2 × 7

Thus, Height of image, h_{2} = – 14 cm

Thus, the height or size of the image is 14 cm. The minus sign shows that this height is in the downward direction, that is, the image is formed below the axis. Thus, the image is real and inverted.

**Example Problem 3.**

The magnification produced by a spherical lens is +2.5. What is the :

(a) nature of image ?

(b) nature of lens ?

**Answer:**

(a) When the magnification is positive, then the image is virtual and erect. In this case, the magnification has a positive sign, so the nature of image is virtual and erect.

(b) The value of magnification given here is 2.5 (which is more than 1). So, the image is larger than the object or magnified. A virtual, erect and magnified image can be formed only by a convex lens, therefore, the nature of lens is convex.