Lenses 

A lens is a piece of transparent glass bound by spherical surfaces (see figure).       

1.0Convex Lens

A convex lens is a piece of a transparent glass bound by two bulging out surfaces. It is thicker at the middle and thinner at the edges. It is also called ‘double convex lens’ or ‘biconvex lens’.

A convex lens is a ‘converging lens’ because the light rays after refraction through it, converge to meet at a single point [see figure (a)].

2.0Concave Lens

A concave lens is a piece of transparent glass bound by two bent-in
(or bulging-in) surfaces. This is thin in the middle but thicker at the edges. It is also called ‘double concave’ or ‘biconcave lens’.

A concave lens is a ‘diverging lens’ because the parallel beam of light rays after refraction through it, appear to diverge from a single point [see figure (b)].

Convex lenses are used as magnifiers in simple microscope, compound microscope, telescope, etc. Convex lenses are also used to correct eye defect ‘hypermetropia’ or ‘long sightedness’.

3.0Basic Terms Used in Lenses 

Optical centre :  The middle point of a spherical lens is called its ‘optical centre’.

Principal axis : The line passing through the optical centre of the lens which is perpendicular to both the faces of the lens is called its ‘principal axis’.

Principal focus : The point on the principal axis of a lens where all the rays parallel to principal axis converge or appear to diverge from after refraction is called ‘principal focus’.

Focal length : The distance between the focus and the optical centre of a lens is called its ‘focal length’.

A convex / concave lens has two principal foci F₁ and F₂. F₁ is towards left of the lens and from this region, light rays are incident on the lens. F₂ is towards right of the lens and in this region, light rays are emergent after refraction.

A lens, either a convex lens or a concave lens, has two spherical surfaces. Each of these surfaces forms a part of a sphere. The centres of these spheres are called centres of curvature of the lens.  Since there are two centres of curvature, we represent them as C₁ and C₂. An imaginary straight line passing through the two centres of curvature of a lens is called its principal axis.

4.0Rules to Obtain Images in Spherical Lenses

Convex lens

(1) A ray from the object parallel to the principal axis, after refraction, passes through the second principal focus F₂ [see figure (a)].

(2) A ray of light passing through the first principal focus in a convex lens, emerges parallel to the principal axis after refraction [see figure (b)].

(3) A ray of light passing through the optical centre of the lens, emerges without any deviation after refraction [see figure (c)].

Concave lens

(1) A ray from the object parallel to the principal axis, after refraction, appears to diverge from the first principal focus F₁ in a concave lens [see figure (a)].

(2) A ray of light appearing to meet at second principal focus F₂ in a concave lens, emerges parallel to the principal axis after refraction [see figure (b)].

(3) A ray of light passing through the optical centre of the lens, emerges without any deviation after refraction [see figure (c)].

Image formation by a concave lens

The image formed by a concave lens is always on the same side as the object and it is always virtual and erect. Also, the size of an image is always diminished, that is, its size is always smaller than that of the object.

Image formation by a concave lens 

5.0Uses of lens

Uses of Convex Lens

1. Convex lens are used in microscopes and magnifying glasses. 

2. Convex lens are used as a camera lens in cameras.

3. Convex lens is used in the correction of hypermetropia. 

Uses of Concave Lenses 

1. Concave lenses are used in a spyhole in doors. 

2. Concave lenses are used in binoculars.

3. Concave lens is used in the correction of myopia.

Activity related to converging of a sun rays after passing through a convex lens

1. Hold a convex lens in your hand. Direct it towards the Sun. Focus the light from the Sun on a sheet of paper (see figure). Adjust the distance between the paper. Adjust the distance between the paper and the lens by moving the lens towards or away from the paper to get a sharp bright image of the Sun.

2. When the paper and the lens are held in the same position for some time, the paper will start burning (see figure).

Reason : The convex lens converges the Sun rays at a point which increases the intensity of light at that point. This increases the temperature of the paper which causes the burning of the paper.

Sign conventions for lens

(1) Optical centre ‘O’ is taken as origin, all the distances along XX’ axis are measured from ‘O’.

(2) Distance along direction of light is considered positive.

(3) Distance measured opposite to the direction of light is considered negative.

(4) Along YY’, the height of object is always taken above principal axis and it is considered positive. For image, if image is virtual and erect (above principal axis), the height is positive. If image is real and inverted (below principal axis), the height is negative. 

6.0Formulae Related to Lenses

Lens Formula

In a lens, the distance of the object from its optical centre is called the object distance (u). The distance of the image from the optical centre of the lens is called the image distance (v). 

The relationship between object distance (u), the image distance (v) and the focal length (f) is given by lens formula which is as given below,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Magnification Formula

The ratio of height of image (h₂) to the height of object (h₁) is called ‘magnification’ or ‘linear magnification’.

$m=\frac{h_2}{h_1}$

The magnification (m) is also related to the object distance (u) and image distance (v). It can be expressed as:

$m=\frac{h_2}{h_1}=+\frac{v}{u}$

Also, magnification can be further expressed as,

$m=\frac{f}{f+u}=\frac{f-v}{f}$

A negative sign in the value of the magnification indicates that the image is real and inverted. A positive sign in the value of the magnification indicates that the image is virtual and erect.

For convex lens, ‘m’ can be +ve as well as –ve. Also, |m| can be >1 or <1 or = 1. For concave lens, ‘m’ is always +ve and |m|<1.

7.0Power of Lens

A lens of short focal length bends the light rays more, through large angles.

The power of a lens is a measure of the degree of convergence or divergence of light rays falling on it

The power of a lens is defined as the ‘reciprocal of its focal length’.        

$P=\frac{1}{f}$

Unit of power of lens :  Diopter (D) 1 diopter = 1 m^–1

The power of convex lens (converging lens) is ‘positive’ and the power of concave lens (diverging lens) is ‘negative’.

Some important points related to lenses

(a) Convex lens

(1) u = always negative.

(2) v = –ve, when object is placed between optical centre (O) and focus (F₁).

v = +ve, for all other possible cases (real and inverted).

(3) f = +ve, power (P) = +ve.

(b) Concave lens

(1) u = always negative.

(2) v =  always –ve (virtual and erect).

(3) f = –ve, power (P) = –ve.

(4) Image always diminished.

For both concave mirror as well as concave lens, focal length ‘f’ is negative. Similarly, for convex mirror and convex lens, focal length ‘f’ is positive.

● Two lenses in contact (lens combination)

(a) Net power of the combination (P) : P = P₁ + P₂

(b) Net or effective focal length of the combination (f) : $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$

(c) Total magnification of the combination (m) = m₁ × m₂

If two thin lenses of equal focal length but of opposite nature are put in contact, the equivalent focal length of combination will be   

$\frac{1}{f_e}=\frac{1}{+f}+\frac{1}{-f}=0\text{i.e.,}{f}_e=\infty .$

Thus, power, $P=\frac{1}{f_e}=\frac{1}{\infty }=0.$

Solved Example

1. 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 =?

Lens equation,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{(-10)}-\frac{1}{u}=\frac{1}{(-15)}$

$\frac{1}{u}=\frac{1}{15}-\frac{1}{10}=\frac{2-3}{30}=\frac{-1}{30}$

$u=-30\text{cm}$

$Magnification,m=\frac{v}{u}=\frac{(-10)}{(-30)}=\frac{1}{3}=+0.33$

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

2. 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

Given, height of the object h₁ = + 2.0 cm ; focal length f = + 10 cm ; object distance u = –15 cm; image distance v =? ; height of the image h₂ =?

Lens equation,

$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \frac{1}{v}-\frac{1}{(-15)}=\frac{1}{(+10)} \end{gathered}$$

$\frac{1}{v}=\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{+1}{30}$

$v=+30\text{cm}$

Magnification, $m=\frac{h_2}{h_1}=\frac{v}{u}$

$h_2=\frac{v}{u}\times h_1=\frac{+30}{-15}\times (+2)=-4\text{cm}$

The negative sign of h₂ shows that the image is inverted and real. A real, inverted image, 4 cm tall, is formed at a distance of 30 cm on the other side of the lens. 

Also, magnification, $m=\frac{v}{u}=+30-15=-2$

3. A magnifying lens has a focal length of10 cm. 

(a) Where should the object be placed if the image is to be 30 cm from the lens? 

(b) What will be the magnification?  

Solution

(a) In case of magnifying lens, the lens is a converging lens i.e. a convex lens. While using it as a magnifying lens, the image is virtual, erect, and magnified. 

Here, f = + 10 cm ; v = – 30 cm (negative sign of v is taken because in case of lens, virtual image is formed on the left side)

By lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{(-30)}-\frac{1}{u}=\frac{1}{(+10)}$

$\frac{1}{u}=-\frac{1}{30}-\frac{1}{10}=\frac{-1-3}{30}=\frac{-4}{30}=\frac{-2}{15}$

$u=-7.5\text{cm}.$

So the object must be placed in front of lens at a distance of 7.5 cm from it.

(b) $m=+\frac{v}{u}=\frac{-30}{-7.5}=+4$

Thus, image is virtual, erect and four times the size of object.

4. An object 25 cm high is placed in front of a convex lens of focal length 30 cm. If the height of real image formed is 50 cm, find the distance between the object and the image. 

Solution

$h_1=25\text{cm},f=+30\text{cm},h_2=-50\text{cm}$

$m=\frac{h_2}{h_1}=\frac{-50}{25}=-2$

$Also,m=+\frac{v}{u}$

$-2=+\frac{v}{u}\text{or}v=-2u$

$Now,\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{(-2u)}-\frac{1}{u}=\frac{1}{(+30)}$

$\frac{-1}{2u}-\frac{1}{u}=\frac{1}{30}$

$\frac{-3}{2u}=\frac{1}{30}$

$u=-45\text{cm}$

$\therefore v=-2u=-2(-45)=+90\text{cm}$

Since object and image are on opposite sides of lens, the distance between object and image d = IuI + IvI = 45 + 90 = 135 cm

5. A needle placed at 45 cm from a lens, forms an image on a screen placed at 90 cm on other side of the lens. Identify the type of lens and determine its focal length. What is the size of the image, if the size of the needle is 5 cm?

Solution

Since the image is formed on the other side of the lens, it is a real image. Such real image can only be formed by a converging lens or convex lens.

Here, u = – 45 cm, v = + 90 cm, f =?, h₂ =?, h₁ = 5 cm, 

By lens equation,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{(+90)}-\frac{1}{(-45)}=\frac{1}{f}$

$\frac{1}{90}+\frac{1}{45}=\frac{1}{f}$

$\frac{1}{f}=\frac{1+2}{90}=\frac{3}{90}$

$f=+30\text{cm}$

Again by calculation also, we got positive value of f i.e., the lens is a converging lens or convex lens.

$Now,m=\frac{h_2}{h_1}=+\frac{v}{u}$

$\frac{h_2}{5}=\frac{90}{-45}$

$h_2=-10\text{cm}$

Minus sign indicates that image is real and inverted.

6. An object 8.5 cm tall is placed at 28 cm from a convex lens that has a focal length of 12 cm. Find the size and location of the image. Draw the ray diagram.

Solution

$\text{Given},h_1=8.5\text{cm};u=-28\text{cm};f=+12\text{cm};v=?;h_2=?$

$\text{By lens equation},\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}-\frac{1}{(-28)}=\frac{1}{(+12)}$

$\frac{1}{v}+\frac{1}{28}=\frac{1}{12}$

$\frac{1}{v}=\frac{1}{12}-\frac{1}{28}$

$\frac{1}{v}=\frac{7}{84}-\frac{3}{84}=\frac{4}{84}=\frac{1}{21}$

$v=+21\text{cm}$

$m=\frac{h_2}{h_1}=+\frac{v}{u}\text{or}\frac{h_2}{8.5}=-\frac{21}{-28}$

$h_2=-6.375\text{cm}$

The image is real, inverted and diminished.

8.0Also Read