# Math in the thin lens equation

Why isn't 1/f=1/do+1/di just the mathematical equivalent to f=do+di? Can't you raise all terms in the equation to the power of negative 1 to get the latter equation? The maths of reciprocals is confusing me lol.

## Answers and Replies

That is not the lens eqn.
The lens equation is :- 1/f = 1/v - 1/u
Where v:- image distance with sign convention
u:- object distance with sign convention
just take reciprocal of whole eqn

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

## f = \frac{1}{\frac{(u-v)}{uv}} ##

## f= \frac{uv}{u-v} ##

Clearly it is self explanatory.

Kuzon
That is not the lens eqn.
The lens equation is :- 1/f = 1/v - 1/u

Kuzon has written a perfectly acceptable version of the lens equation, one that is commonly used.

Kuzon: raising to a power does not follow a distributive law as does multiplication. So we have a(b + c) = ab + ac but (b + c)2 ≠ b2 + c2.

Kuzon and Sahil Kukreja
Kuzon has written a perfectly acceptable version of the lens equation, one that is commonly used.

Kuzon: raising to a power does not follow a distributive law as does multiplication. So we have a(b + c) = ab + ac but (b + c)2 ≠ b2 + c2.

Sorry, I did not know that version of lens equation; as in our country the lens equation version used is different.

I have acquired the same as @Sahil Kukreja . The relation is about their reciprocals, and cannot directly be changed to what @Kuzon wrote.

Sahil Kukreja