1. ## sovle equation

If is the focal length of a lens and is the distance of an object from the lens and is the distance of the image from the lens then
(a) Solve this equation for the distance of the image in terms of the other two quantities. Simplify your answer and put the denominator of the simplified fraction into the answer box (notice the numerator) (b) If the focal length is and the distance of the object from the lens is what is the distance of the image from the lens? (c) If the focal length is and the distance of the object from the lens is very large, for example what is the distance of the image from the lens approximately? (this explains the name focal length)

2. Hello, lilikoipssn!

If $\displaystyle f$ is the focal length of a lens
and $\displaystyle u$ is the distance of an object from the lens
and $\displaystyle v$ is the distance of the image from the lens,

then: . $\displaystyle \frac{1}{u} + \frac{1}{v}\;=\;\frac{1}{f}$

(a) Solve this equation for $\displaystyle v.$

Muliply by $\displaystyle uvf\!:\quad fv + fu \:=\:uv\quad\Rightarrow\quad uv-fv \:=\:fu$

Factor: .$\displaystyle (u-f)v \:=\:fu \quad\Rightarrow\quad\boxed{ v \:=\:\frac{fu}{u-f}}$

(b) If $\displaystyle f = 8$ cm and $\displaystyle u = 38$ cm, find $\displaystyle v.$

We have: .$\displaystyle v \;=\;\frac{(8)(38)}{38-8} \;=\;\frac{304}{30} \;=\;\boxed{\frac{152}{15}\text{ cm}}$

(c) If $\displaystyle f = 8$ cm, and $\displaystyle u$ is very large, for example, 1,000,000 cm
find $\displaystyle v$ approximately. (This explains the name focal length)

We have: . $\displaystyle v \;=\;\frac{(8)(1,\!000,\!000)}{1,\!000,\!000 -8} \;\approx\;\frac{8,\!000,\!000}{1,\!000,\!000} \;=\;8$