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## Magnification Formula for lens in terms of focal length

The lens formula for a concave lens is the same as that for a convex lens, which is:

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

The magnification formulae for a concave lens are also just the same as that for convex lens, which are:

m = \(\frac{h_2}{h_1}\) and m = \(\frac{v}{u}\)

### Numerical Problems Based on Concave Lenses

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

**Example Problem 1**.

An object is placed at a distance of 50 cm from a concave lens of focal length 20 cm. Find the nature and position of the image.

**Solution.**

First of all we will find out the position of image which is given by the image distance v.

Here, Object distance, u = -50 cm (It is to the left of lens)

Image distance, v = ? (To be calculated)

Focal length, f = -20 cm (It is a concave lens)

Putting these values in the lens formula :

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

we get : \(\frac{1}{v}\) = \(\frac{1}{-50}\) = \(\frac{1}{-20}\)

or \(\frac{1}{v}\) + \(\frac{1}{50}\) = –\(\frac{1}{20}\)

or \(\frac{1}{v}\) = –\(\frac{1}{20}\) – \(\frac{1}{50}\)

or \(\frac{1}{v}\) = \(\frac{-5-2}{100}\)

or \(\frac{1}{v}\) = –\(\frac{100}{7}\)

So, Image distance, v = – 14.3 cm

Thus, the image is formed at a distance of 14.3 cm from the concave lens. The minus sign for image distance shows that the image is formed on the left side of the concave lens. We know that a concave lens always forms a virtual and erect image, so the nature of image is virtual and erect.

**Example Problem 2.**

An object placed 50 cm from a lens produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of image. Calculate focal length of the lens and magnification produced.

**Solution.**

First of all we will find out the focal length of the lens. We know that the object is always placed in front of the lens on the left side, so the object distance is always taken as negative. Here the image is also formed in front of the lens on the left side, so the image distance will also be negative. Thus,

Object distance, u = – 50 cm (To the left of lens)

Image distance, v = – 10 cm (To the left of lens)

Focal length, f = ? (To be calculated)

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

we get, \(\frac{1}{-10}\) – \(\frac{1}{-50}\) = \(\frac{1}{f}\)

\(\frac{1}{-10}\) + \(\frac{1}{50}\) = \(\frac{1}{f}\)

\(\frac{-5+1}{50}\) = \(\frac{1}{f}\)

\(-\frac{4}{50}\) = \(\frac{1}{f}\)

f = \(-\frac{50}{4}\)

So, Focal length, f = – 12.5 cm

The minus sign for focal length shows that it is a concave lens. Please draw the ray diagram yourself. We will now calculate the magnification produced by concave lens. We know that for a lens :

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

Here, Image distance, v = -10 cm

And, Object distance, u = – 50 cm

So, m = \(\frac{-10}{-50}\)

m = +\(+\frac{1}{5}\)

m = + 0.2

Thus, the magnification produced by this concave lens is + 0.2. Since the value of magnification is less than 1 (it is 0.2), therefore, the image is smaller than the object (or diminished). The plus sign for the magnification shows that the image is virtual and erect.

**Example Problem 3.**

The magnification produced by a spherical lens is + 0.75. What is the :

(a) nature of image ?

(b) nature of lens ?

**Answer:**

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

(b) The value of magnification given here is 0.75 (which is less than 1), so the image is smaller than the object or diminished. A virtual, erect and diminished image can be formed only by a concave lens, so the nature of lens is concave.