[Physics Class Notes] on Derivation of Lens Formula Pdf for Exam

In physics, a lens is defined as a device that either focuses or disperses the light beam falling on it using refraction. Based on this concept, the lenses are classified into two types – converging lenses that concentrate parallel rays of light falling on them and diverging lenses that cause parallel rays of light to spread out. No matter whether converging or diverging, both types of lenses are marvels of optical physics and used to create a sharp, magnified, and clear image of the object placed on one side of them. Although the principal purpose of all the lenses is to magnify or we can say make images appear larger than their actual size, still there is a remarkable difference in the images formed by them. For instance, an image formed by a converging lens differs from one formed by a diverging lens. There are several aspects like shapes, physical dimensions, etc., of a lens that impact the behaviour of a light beam falling on it, and also the characteristics of the image formed. To understand the physics of the concept of lenses and images formed by them, we need to know about lens formula. It is a key term around which our optical physics often revolves.

 

What is Lens Formula?

Based on the physics concept stating that lenses are formed by coupling two spherical surfaces together, lenses are of two types:

Characteristics of images created by these lenses differ depending on the aspects of lenses and object’s distance from these lenses. It is where the lens formula comes into the action. As per optical physics, lens formula relates the distance of an object (u), the distance of an image (v), and the focal length (f) of the lens. Applicable for both the convex and concave lenses, the lens formula is given as:

1/v – 1/u = 1/f  

Where, 

v = Distance of image formed from the optical center of the lens. 

u = Distance of object from the optical center of the lens.

f = focal length of the lens.  

 

Lens Formula Derivation

Convex Lens

Consider a convex lens with O be the optical centre, and F be the principal focus with focal length f. Now, let AB be the object kept perpendicular to the principal axis and at a distance beyond the focal length.                      

As the object is perpendicular to the principal axis, the image will also be perpendicular to the principal axis. To find out the location of the image formed, draw a perpendicular from point A’ to point B’ on the principal axis. On doing this, we can see that the ΔABO and ΔA’B’O are similar as shown in the figure. 

So, A′B′/AB = OB’/OB (as ΔABO and ΔA’B’O are similar) … (1)

Also, ΔA’B’F and ΔOCF are similar

So, A’B’/OC = FB’/OF 

But, OC = AB

Therefore, A’B’/AB = FB’/OF 

From the above equations, we get:

OB’/OB = FB’/OF = (OB’ – OF)/ OF

Now, by using the sign convention, OB = -u, OB’ = v, and OF = f, we can say:

v/-u = (v – f) /f

=> vf = – uv + uf or uv = uf – vf

Diving both sides by uvf, we will have:

uv / uvf = (uf / uvf) – (vf/ uvf)

Hence, 1/f = 1/v – 1/u 

This is the required lens formula.

 

Concave Lens

Let AB be the object perpendicular to the principal axis and at a distance more than the focal length (f) of the convex lens. We will see that the image A’B’ is erect, virtual, and formed on the same side as the object.                   

Now, from the figure, we can consider that:

OF1 is the focal length (f),

OA is the object distance (u),

OA is the image distance (v),   

and ΔOAB and ΔOA’B’ are similar

 ∵ Angle BAO = Angle B’A’O = 900, vertex O is common for both the triangles

So, Angle AOB = Angle A’OB’

Therefore, Angle ABO = Angle A’B’O

And, A’B’ / AB = OA’ / OA                                              … (1)

Again, ΔOCF1 and ΔF1A’B’ are similar

So, A’B’/ OC = A’F1/ OF1

But from the diagram, we can see that OC = AB

A’B’ / AB = A’F1/ OF1 = (OF1 – OA’)/ OF1

A’B’ / AB = (OF1 – OA’)/ OF1                                    … (2)

From equation (1) and equation (2), we get

OA’ / OA = (OF1 – OA’)/ OF1       

– v / -u = (-f – – v)/ -f                                                                      

v/ u = (-f + v)/ -f   

– vf = – uf  + uv                        … (3)                                                                

Dividing equation (3) by uvf

– 1/u = – 1/v + 1/f

Hence, 1/f = 1/v – 1/u 

This is the required lens formula.

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