**BASIC LEVEL** **TOPIC-WISE ****ONLINE TEST SERIES**

# Welcome to the Living World

A Treasure of Knowledge for Biology Lovers

## Friday, September 30, 2022

### Basic 10 | Body Fluids and Circulation | Topic 1: Blood and Lymph

**BASIC LEVEL** **TOPIC-WISE ****ONLINE TEST SERIES**

## Wednesday, September 28, 2022

### 10. Light - Reflection and Refraction | Class 10 CBSE | Web Notes | Part 9: Lens Formula & Magnification, Power of Lenses

# 10. LIGHT – REFLECTION AND REFRACTION

**Lens Formula and Magnification**

Formula for spherical lenses gives the relationship between **object
distance ( u), image-distance (v)** and the

**focal length (**The lens formula is expressed as

*f*).The lens formula is
general and is valid in all situations for any spherical lens.

The **magnification ( m)**
produced by a lens is similar to that for spherical mirrors. It is the

**ratio**of the

**height of**the

**image (**and

*h’*)**the**

**height**of the

**object (**

*h*).Magnification is also
related to the **object-distance u,**

*and the*

**image-distance**

*v*.Magnification *(m*)
= *h*′/*h *= *v*/*u*

**Problem: **A concave lens has focal
length of 15 cm. At what distance should the object from the lens be placed so
that it forms an image at 10 cm from the lens? Also, find the magnification
produced by the lens.

**Solution:**** **A concave lens always forms a virtual, erect image on the same
side of the object.

Image-distance
*v *= –10 cm

Focal
length *f *= –15 cm

Object-distance
*u *= ?

or, *u *= __– 30 cm__

Thus,
the object-distance is **30 cm.**

Magnification
*m *= *v/u*

The
positive sign shows that the image is erect and virtual. The image is one-third
of the size of the object.

**Problem: **A 2.0 cm tall object is placed
perpendicular to the principal axis of a convex lens of focal length 10 cm. The
distance of the object from the lens is 15 cm. Find the nature, position and
size of the image. Also find its magnification.

**Solution****
**Height of the object

*h*= + 2.0 cm

Focal length *f *= +
10 cm*
*object-distance

*u*= –15 cm

*Image-distance*

*v*=?

Height of the image *h*′=?

or, *v *= __+ 30 cm__

The
positive sign of *v *shows that the image is formed at a distance of 30 cm
on the other side of the optical centre. The image is **real** and **inverted.**

or, *h*′ = *h *(*v*/*u*)

Height
of the image, *h*′ = (2.0) (+30/–15) = __–
4.0 cm__

Magnification
*m *= *v/u*

The negative signs of *m *and *h*′ show that the image
is **inverted** and **real.** It is formed below the principal axis.
Thus, a real, inverted image, 4 cm tall, is formed at a distance of 30 cm on
the other side of the lens. The image is **two times enlarged.**

**Power of a Lens**

The ability of a lens to converge or diverge light rays depends on its focal length.

A **convex lens** of **short focal length** bends the
light rays through large angles and **focus closer to the** **optical
centre. **

A **concave lens** of **very
short focal length** causes **higher divergence** than the one with longer
focal length.

The degree of convergence
or divergence of light rays by a lens is expressed as its **power.**

The **power of a lens ( P)**
is defined as the

**reciprocal of its focal length (**

*f*).The SI unit of power of a
lens is **dioptre (D).**

**1 dioptre** is the power of a lens whose **focal length is 1 metre** **(1D
= 1m ^{–1}).**

The *power** of a convex lens is positive
and that of a concave lens is negative.*

Opticians prescribe
corrective lenses indicating their powers. It is more convenient to use powers
instead of focal lengths. E.g.

o A lens of power **+2.0 D **has a
focal length of **+0.50 m. **The lens is **convex.**

o A lens of power **–2.5 D **has a focal length of **–0.40 m.**
The lens is **concave.**

In many optical
instruments, several lenses are combined to increase magnification and
sharpness of the image. The **net power ( P)** of the lenses placed in
contact is the algebraic sum of the individual powers

*P*,

_{1}*P*,

_{2}*P*, … (

_{3}*P*=

*P*+

_{1}*P*+

_{2}*P*+ …).

_{3}During **eye-testing,**
an optician puts several corrective lenses of known power inside the testing
spectacles’ frame. Thus the required power of the lens is calculated by
algebraic addition. E.g., two lenses of power + 2.0 D & + 0.25 D is
equivalent to a single lens of power + 2.25 D.

**lens systems**to minimise certain defects in images produced by a single lens. Such a lens system is commonly used in the design of lenses of camera, microscopes and telescopes.