Convex Lens

It  is a transparent optical device that converges light rays passing through it to a single point called the focal point. It is thicker at the center than at the edges and is commonly used to form real or magnified images. Convex lenses are found in everyday objects like eyeglasses, magnifying glasses, and cameras. They come in different types, including plano-convex and double-convex lenses, and are essential in optical systems due to their ability to focus light.

1.0Definition of Convex Lens

• A lens is a portion of a transparent material with two refracting surfaces such that at least one is curved with the refractive index of its material being different from that of the surroundings.
• A thin spherical lens with refractive index greater than that of surroundings behaves as a convergent or convex lens, i.e., converges parallel rays if its central (i.e. paraxial) portion is thicker than marginal one.

2.0Types of Convex Lens

3.0Principal Focus of Convex Lens

Principal Focus: A Lens has two focal points.

First Focus: First focal point is an object point on the principal axis corresponding to which the image is formed at infinity.                  

Second Focus: Second focal point is an image point on the principal axis corresponding to which object lies at infinity.   

4.0Rules For Image Formation In Convex Lens

• A ray passing through the optical centre proceeds without deviation through the lens.
•  A ray passing through first focus or directed towards it, becomes parallel to the principal axis after refraction from the lens.
• A ray passing parallel to the principal axis passes or appears to pass through F after refraction through the lens.

5.0Image Formation In Convex Lens

6.0Image formation in Convex Lens

• Position of Object- Placed at Infinity 
• Position of Image-Formed at f
• Nature-Real, Inverted
• Size-Highly Diminished                        

• Position of Object- Placed at -∞ and 2f
• Position of Image-Formed in between 2f and f
• Nature-Real,Inverted
• Size- Diminished            

• Position of Object- Placed at  -2f
• Position of Image-Formed at 2f
• Nature-Real,Inverted
• Size- Equal in size

• Position of Object- Placed in between -2f and -f
• Position of Image-In between 2f and ∞
• Nature-Real,Inverted
• Size- Enlarged      

• Position of Object- Placed in between -2f and -f ( very nearer to -f)
• Position of Image-Formed At ∞
• Nature-Real,Inverted
• Size- Enlarged      

• Position of Object-  Object is Placed in between -f and optical centre  ( very nearer to -f)
• Position of Image-Formed At - ∞
• Nature-Virtual,Erect
• Size- Highly Enlarged                 

• Position of Object- Object is Placed in between -f and optical centre
• Position of Image-In between  - ∞ and Optical Centre
• Nature-Virtual,Erect
• Size-  Enlarged

7.0Lens Formula For Convex Lens             

Object AB is positioned between the optical centre and and focus of the convex lens.Image A’B’ formed by convex lens is virtual,erect and magnified.

$\Delta A^{\prime }{B}^{\prime }O\text{ and }ABO\text{ are similar}$

$\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{B^{\prime }O}{B^{\prime }O}$……….(1)

$\Delta A^{\prime }{B}^{\prime }F\text{ and }MOF$are similar

$\frac{A^{\prime }{B}^{\prime }}{MO}=\frac{BF}{OF}$

MO=AB therefore

$\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{BF}{OF}$……….(2)

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

$vf=uf-uv\Rightarrow uv=uf-vf$

Dividing both sides by uvf we get

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

8.0Magnification for Convex Lens

• It is defined as the ratio of the size of image formed by the lens to the size of the object.

$m=\frac{\text{Size of Image}}{\text{Size of Object}}=\frac{h_2}{h_1}=\frac{v}{u}$

• Linear Magnification in terms of u and f

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

Multiplying both sides by u

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

$m=\frac{1}{u}=\frac{f}{f+u}$

• Linear Magnification in terms of v and f

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

Multiplying both sides by v

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

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

Example-1.An object and a screen are positioned 90 cm apart. What is the required focal length and type of lens needed to form a distinct image on the screen, two times the the size of the object?

Solution:Since the image is real, a convex lens should be placed between the object and the screen.

u=-x    

v=90-x

$m=\frac{v}{u}=-2$

$\frac{90-x}{-x}=-2\Rightarrow 90-x=2x\Rightarrow x=30$

u=-30 cm, v=+60 cm

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{60}-\frac{1}{-30}=\frac{3}{60}=\frac{1}{20}\Rightarrow f=20\text{ cm}$

Q-2.An image formed by a convex lens is upright and four times the size of the object. Given the focal length of the lens is 20 cm,Find the object and image distances.

Solution:

To calculate u

$m=\frac{f}{f+u}\Rightarrow 4=\frac{20}{u+20}\Rightarrow u=-15\text{ cm}$

To calculate v

$m=\frac{f-v}{f}\Rightarrow 4=\frac{20-v}{20}\Rightarrow v=-60\text{ cm}$