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Wize University Physics Textbook (Master) > Geometric Optics

Lenses Equation

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Lenses Equation



The distance of the image in relation to the object is described by the mirrors & lenses equation:


 1f=1di+1do \boxed{ \ \dfrac{1}{f}=\dfrac{1}{d_i}+\dfrac{1}{d_o} \ }

  • dod_o is the distance between the object and the lens
  • did_i is the distance between the image and the lens
  • ff is the focal length of the lens

The magnification of the lens is given by:

 m=−dido=hiho \boxed{ \ m=-\frac{d_i}{d_o}=\frac{h_i}{h_o} \ }
  • mm is the magnification
  • hoh_o is the height of the object
  • hih_i is the height of the image


Note: you might see different notations for the distances ( did_i and dod_o, ii and oo, pp and qq ).



Wize Concept
  • ff is positive for convex lenses, and negative for concave lenses
  • dod_o and hoh_o are always positive
  • did_i is positive if the image is on the other side of the lens, and negative if it's on the same side
  • hih_i is positive if the image is upright, and negative if inverted
  • mm is positive if the image is upright, and negative if inverted
  • real images are inverted, virtual images are upright


Exam Tip
Using these equations we can answer the following questions:
  1. Where is the image?
  2. Is it real or virtual?
  3. Is it upright or inverted?
  4. How big is the image (relative to the object)?

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Example: Converging vs Diverging Lens

A lens is used to view a small shell found on the beach.

a) If the shell is magnified by a factor of 2.252.25 when the shell is held 3.23.2 cm from the lens, what is the focal length? What kind of lens is it? Draw a ray diagram.

b) If the shell is magnified by a factor of 0.80.8 when the shell is held 3.23.2 cm from the lens, what is the focal length? What kind of lens is it? Draw a ray diagram.


Part a)


Let's use the magnification equation to get the image distance:

m=−dido   →   di=−m dom=-\dfrac{d_i}{d_o} \ \ \ \to \ \ \ d_i=-m \ d_o

Putting this into the lens equation we get:

1f=1di+1do\dfrac{1}{f}=\dfrac{1}{d_i}+\dfrac{1}{d_o}

1f=−1m do+1do\dfrac{1}{f}=-\dfrac{1}{m\ d_o}+\dfrac{1}{d_o}

= 1do(1−1m)=\ \dfrac{1}{d_o}\bigg(1-\dfrac{1}{m}\bigg)

Using m=2.25m=2.25 and do=3.2d_o=3.2 we get:

1f= 1do(1−1m)\dfrac{1}{f}=\ \dfrac{1}{d_o}\bigg(1-\dfrac{1}{m}\bigg)

= 13.2(1−12.25)=\ \dfrac{1}{3.2}\bigg(1-\dfrac{1}{2.25}\bigg)

Taking the reciprocal of both sides, the focal length is:

f=5.76f=5.76 (cm)

Since the focal length is positive, the lens is a convex (converging) lens.


Part b)


Using the same equation as before but with m=0.8m=0.8 and do=3.2d_o=3.2 we get:

1f= 1do(1−1m)\dfrac{1}{f}=\ \dfrac{1}{d_o}\bigg(1-\dfrac{1}{m}\bigg)

= 13.2(1−10.8)=\ \dfrac{1}{3.2}\bigg(1-\dfrac{1}{0.8}\bigg)

Taking the reciprocal of both sides, the focal length is:

f=−12.8f=-12.8 (cm)

Since the focal length is negative, the lens is a concave (diverging) lens.


An object is placed 25.025.0 cm in front of a converging lens of focal length 10.010.0 cm. A second converging lens of the same focal length is placed 50.050.0 cm behind the first. Where is the final image located? What is the magnification of the final image? Draw a ray diagram illustrating the situation.

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Location of final image: