Ray Diagram for Converging Lenses
Ray 1 : rays (photons) from a point on the object that travel parallel to the axis will hit the lens and head towards F.   Ray 2 : A ray that passes through F' will refract into a ray parallel to the axis (since a photon coming from the right along that exact path would refract that way).   Ray 3 : A ray that passes through the center of the lens (the vertex) will ultimately exit at the same angle it came in with, appearing to just keep doing in a straight line.
Example : Suppose we have a converging lens with a 20 cm focal length and we place an object 30 cm to the left of this lens. Where will the image form? Is it real or virtual? Upright or inverted? Larger or smaller than the object?   Example : Suppose we have a converging lens with f=20 cm and we place an object 10 cm to the left of this lens. Where will the image form? Is it real or virtual? Upright or inverted? Larger or smaller than the object?

 

Ray Diagram for Diverging Lenses
Ray 1 : rays (photons) from a point on the object that travel parallel to the axis will hit the lens and head off to the right, but appear to have come from F.   Ray 2 : If we draw a ray aimed at F' over on the right, it will refract into a ray parallel to the axis (since a photon coming from the right along that exact path would refract that way).   Ray 3 : A ray that passes through the center of the lens (the vertex) will ultimately exit at the same angle it came in with, appearing to just keep doing in a straight line.
Example : Suppose we have a diverging lens with a 20 cm focal length and we place an object 30 cm to the left of this lens. Where will the image form? Is it real or virtual? Upright or inverted? Larger or smaller than the object?   Example : Suppose we have a converging lens with f=20 cm and we place an object 10 cm to the left of this lens. Where will the image form? Is it real or virtual? Upright or inverted? Larger or smaller than the object?

 

Thin Lens Equation; Magnification
Suppose we have a converging lens with an object outside the focal point. The two rays shown create two pairs of similar triangles: Triangle FI'I is similar to triangle FBA Triangle OAO' is similar to triangle IAI' We can use these to derive: \[ m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \hspace{2em} Magnification \] \[ \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f} \hspace{2em} Lens~Equation\]   (Exactly the same equation we used for mirrors.)
What about diverging lenses?
Suppose we have a diverging lens with an object outside the focal point. The two rays shown create two pairs of similar triangles: Triangle IAI' is similar to triangle OAO' Triangle IFI' is similar to triangle AFB Unfortunately these lead to: \[ \frac{1}{d_o} + \frac{1}{(-d_i)} = \frac{1}{(-f)} \]   We can MAKE the original equation work if we adopt a sign convention here. For a diverging lens, the focal point is NOT on the outgoing ray side of the lens (like it was for a converging lens), so let's treat that as a negative value. Also, here the image is NOT on the outgoing ray side, so consider that a negative distance too.

 

Sign Conventions
• The focal length f is positive for converging lenses and negative for diverging lenses.   • The object distance is positive if the object is on the side of the lens from which the light is coming (i.e. the incoming ray side). You'd think this would always be the case, but when we start to deal with multiple lenses, like we did with multiple mirrors, we can end up with a negative object distance.   • The image distance is positive if the image is on the opposite side of the lens where the light is coming (i.e., on the outgoing ray side); if not, the image distance is negative.     di>0 (image on outgoing ray side) implies a REAL image.     di<0 (image NOT on outgoing ray side) implies a VIRTUAL image.   • The height of the image hi is positive if the image is upright (relative to the object) and negative if the image is inverted (relative to the object).   These sign conventions work for mirrors as well (although in that case the INCOMING and OUTGOING ray sides are the same side), giving us a single equation that works for both mirrors and lenses: \[ m = \frac{h_i}{h_o} = -\frac{ d_i }{d_o } \hspace{2em} Magnification \] \[ \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f} \hspace{2em} Lens/Mirror Equation \]
Converging Lens	Diverging Lens

 

Examples
EXAMPLE 1 : Suppose we have a converging lens with f=+20 and place an object 30 cm to the left of this lens. Use a ray diagram to determine where the image will form. Is it real or virtual? Upright or inverted? Larger or smaller than the object? Use the lens equations, which should yield numerical values that are consistent with the ray diagram.
EXAMPLE 2 : Suppose we have the same lens but place an object just 10 cm from the lens.       (This would be the typical 'magnifying glass' geometry.)
EXAMPLE 3 : Suppose we have a diverging lens with a 20 cm focal length and place an object 10 cm to the left of this lens. Use a ray diagram to determine where the image will form. Is it real or virtual? Upright or inverted? Larger or smaller than the object? Use the lens equations, which should yield numerical values that are consistent with the ray diagram.
EXAMPLE 4 : Suppose we have the same diverging lens but place the object 30 cm to the left of the lens.