**Lens :-** A transparent material bound by two surfaces, of which one or both surfaces are spherical is called the lens.

**There are two types of lenses:**

**i. Convex (Converging) lens-** Convex lenses are thick in the middle and thinner at the edges. A convex lens is also known as the converging lens because it converges parallel light rays inwards to a point known as the focal point.

**ii. Concave (Diverging) lens-** Concave lenses are flat in the middle and thicker at the edges. The concave lens is also known as a diverging lens because it diverges the parallel light rays outward from a point knows as the focal point.

**Lens formula:**

If u = object distance, v = image distance and

f = focal length of lens,then the lens formula is given by

According to sign conventions we need to assume the distance as follows,

Object distance, **u= -ve** , Image distance, **v = -ve/+ve** (depends upon the position of image) and Focal length, **f= -ve for concave lens/+ve for convex lens.**

Let’s consider following example to understand this formula more precisely.

**Ex.1) An object is kept in front of converging lens of focal length 20 cm at distance of 10 cm. Find position of image, nature of image.**

**Solution:** Data given ;( using all sign conventions)

Focal length, f= +20 cm

Object distance, u= -10 cm

Image distance=?

Formula:

On substituting the above values, we get

∴ 1/v = 3/20

∴ v = 20/3

∴ v = 6.67 cm

Positive value of ‘v’ indicates that image is real but inverted.

**Magnification by lenses (M):**

Magnification is defined as the ratio of height of image (h_{2}) to the height of object (h_{1}) or the ratio of image distance (v) to object distance (u).

**Ex.2) An object of height 5 cm kept in front of convex lens of focal length 25 cm at distance of 10 cm. Find position of image, nature of image and height of the image.**

**Solution:** Data given ;( using all sign conventions)

Height of object, h_{1}= + 5 cm

Focal length, f= +25 cm

Object distance, u= -10 cm

Image distance=?

Image height=?

Formula:

∴ 1/v = 35/200

v = 200/35

∴ v = 5.7 cm

Positive value of ‘v’ indicates that image is real but inverted.

Now to calculate height of image use 2^{nd} formula

Hence the height of image obtain here will be 2.85 cm.

**(3) The distance of object is 20cm from convex mirror of focal length is 25 cm. Find the position of image?**

__Given:__

F = 15 => convex mirror

Object distance u = -20 cm

Image distance (v) =?

1/f = 1/v + 1/u

1/25 = 1/v + 1/-20

1/25 + 1/20 = 1/v

45/500 = 1/v

V = 500/45 = 11.11

V = 11.1100

The value of V is +ve so image is virtual.

**(4) The radius of curvature is 38 cm. Find the focal length of convex mirror?**

__Given:__

R = 38 cm

Focal length is equal to half of radius of curvature

Focal length = R/2

= 28/2

F = 19 cm

Focal length is 19 cm

**(5) Find the position of image if the focal length of convex lens (f) is 8 cm and object is placed 16 cm from lens.**

__Given:__

f = 8 cm

u = – 16 cm

v =?

1/f = 1/v – 1/u

1/8 = 1/v – 1/16

1/v = 1/8 – 1/16

1/v = 1/16

V = 16 cm

∴ The image is placed 16 cm from the principal axis.

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