How do You Determine the Focal Length of a Lens?
Important Formula
What do the formula symbols mean?
Focal length
$$ f $$ Unit $$ \mathrm{m} $$For example, if the object is 1 meter away from the lens, so \(g = 1 \, \text{m} \) and the sharp image of the object is 2 centimeters away from the lens, so \(b = 0.02 \, \text{m} \). Then the focal length of the lens is: \begin{align} f &~=~ \frac{ 1 \, \text{m} ~\cdot~ 0.02 \, \text{m}}{ 1 \, \text{m} ~+~ 0.02 \, \text{m} } \\\\ &~=~ 0.02 \, \text{m} \end{align}
Thus, the focal point of the lens used is at a distance of 2 centimeters from the lens.
Object distance
$$ g $$ Unit $$ \mathrm{m} $$Image distance
$$ b $$ Unit $$ \mathrm{m} $$Table of contents
To determine the focal length of a lens, you must create a sharp image of an object on a screen - with a lens. Then you measure the distance \( b \) (screen-lens), as well as the distance \( g \) (object-lens) and can then determine the focal length \( f \) with the help of the lens equation:
Or you can focus a far away object (\( g ~\rightarrow~ \infty \): \(1/g ~=~ 0\)), then the image, because of Eq. 1
, lies on the screen exactly at the focal length: \( f ~=~ b \).
Exercises with Solutions
Use this formula eBook if you have problems with physics problems.Exercise #1: Focal Length of a Biconvex Thin Lens
Determine the focal length \(f\) of a biconvex thin lens when the object located 10 centimeters away produces a sharp image at a distance of 1.4 meters.
Solution to Exercise #1
According to the exercise, the biconvex lens is a thin lens; that is: Its thickness is small compared to the two radii of curvature, so you can use the lens equation: \[ \frac{1}{f} ~=~\frac{1}{b} ~+~ \frac{1}{g} \]
So, if you're dealing with a thin lens (whether biconvex, concave, or otherwise) and you know the distance from the lens to the object, as well as from the lens to the sharp image, you also have the focal length \(f\) of the lens. Rearrange the lens equation to solve for the focal length: \[ f ~=~ \frac{b \, g}{b ~+~ g} \]
Substitute the specific values from the problem statement (with \(g\)=10cm=0.1m): \[ f ~=~ \frac{1.4 \text{m} ~\cdot~ 0.1 \text{m} }{1.4 \text{m} ~+~ 0.1 \text{m}} ~=~ 0.093 \text{m} \]