Following graphic illustrates a simple lens model:

where,

h= height of the object

h’= height of the object projected in an image

G and C = focal points

f= focal distance

u= Distance between the object and the focal point

O= Centre of the lens

v= Distance between the centre of the lens and image plane

** Assumptions**

- Lens is very thin
- Optical axis is perpendicular to image plane

**To Prove**

1/f=1/u + 1/v

**Proof**

In ΔAHO, tanα=h/u

In ΔEDO, tanα=h’/v

∴ tanα=h/u=h’/v

⇒ h’/h=v/u ————- (1)

In ΔBOC, tanβ=h/f

In ΔEDC, tanβ=h’/(v-f)

∴ tanβ=h’/(v-f)=h/f

⇒ h’/h=(v-f)/f ———- (2)

Equating (1) and (2),

v/u=(v-f)/f

⇒ v/u=v/f -1

Dividing both sides by v,

1/u=1/f – 1/v

or,

1/f=1/u + 1/v

Hence proved.

**Notes**

h’/h is often referred to as the Magnification factor *M. *If M is negative, the projected image is real but inverted.

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mark tanzi(23:32:13) :Nice proof of the lens law…thank you.

I noticed that the u in your diagram should extend from H to O, not from H to G.

Thanks.

siddahuja(18:12:24) :Thanks..just edited it.

mark tanzi(21:55:38) :Hello Siddahuja, I wrote in on 11/30 regarding changing dimension u to extend from H to O, and you upated accordingly….thank you.

I have a question for you, that has long puzzled me in the derivation of the thin lens equation. In your proof, you state that the tangent of beta = h/f. h is defined as the height of the object, and the diagram shows that h is equal to BO, or half the dimension of the lens as shown. Since the lens dimension BO is typically very small in relation to an object height h (a lens may only be 25mm or smaller…), is this proof assuming that the object height, h, is also very small? Does this proof only apply to object heights h that are no larger than the physical dimension of a lens? The trigonometry is simple enough to understand, but I am having difficulty with the concept of the object height h falling within the dimension of the lens BO. If you would take a moment to explain, it would be greatly appreciated. Thank you very much.

Siddhant Ahuja (Sid)(23:17:39) :Thanks for going over it!

The object can be placed at a distance less than, equal to or greater than the focal distance. It is assumed in the above proof that for a converging lens, incoming light rays that are parallel to the optical axis and parallel to each other converge (refraction) to a common focal point after going through the lens. The object height may exceed the dimensions of the lens, but the distance between the lens and the object will greatly influence the image size (height and width of the image formed on the screen) and quality. Thus, the full dimensions of a large object cannot be seen if it is closer, as opposed to farther away. The drawback will be the ability of the lens to focus the finer details of the object.

mark tanzi(00:27:13) :Thanks Sid.

I re-arranged your Lens Law proof, adapting the math to an object of greater physical size than the lens. May I email you a quick sketch for your review? If so, would you email me your email address, at marktanzi@msn.com. It would be great if you could have a look. Thanks.

mark

muhammad ilyas(22:21:48) :i learn many thing from this lens law

sagar(08:33:04) :thanks. It helped in my class test . Using 3g i got the answer at the right time.

margarita(18:18:39) :kul