Lens formula: 1/v - 1/u = 1/f
For a thin lens, 1/v - 1/u = 1/f. A convex lens with f = 10 cm and an object 15 cm away gives v = +30 cm and m = -2: a real, inverted, enlarged image.
The ray diagram
Drag the object or type values. Signs follow the New Cartesian convention: distances are measured from the pole or optical centre, along the incident light is positive, against it negative, and heights above the axis are positive. So u is negative for a real object.
Convex lens of focal length +10 cm. An object +2 cm tall at u = -15 cm: the image is at v = +30 cm, real, inverted, enlarged.
Show the working
- f = +10 cm, u = -15 cm (the object is in front, so u is negative), h = +2 cm.
- 1/v = 1/f + 1/u = 1/(+10) + 1/(-15) = 5/150 = 1/30
- v = +30 cm: positive, so the image is behind the lens and real.
- m = v/u = (+30)/(-15) = -2: negative, so inverted, and enlarged.
- h′ = m × h = -2 × +2 = -4 cm
- Power P = 1/f = 1/(+0.1 m) = +10 D
Convex lens, f = 10 cm, object at 15 cm
- An object is placed 15 cm from a convex lens of focal length 10 cm. Find the image.
- f = +10 cm, u = -15 cm (the object is in front, so u is negative), h = +2 cm.
- 1/v = 1/f + 1/u = 1/(+10) + 1/(-15) = 5/150 = 1/30
- v = +30 cm: positive, so the image is behind the lens and real.
- m = v/u = (+30)/(-15) = -2: negative, so inverted, and enlarged.
- h′ = m × h = -2 × +2 = -4 cm
- Power P = 1/f = 1/(+0.1 m) = +10 D
Concave lens, f = 15 cm, object at 10 cm
- An object is placed 10 cm from a concave lens of focal length 15 cm. Find the image.
- f = -15 cm, u = -10 cm (the object is in front, so u is negative), h = +2 cm.
- 1/v = 1/f + 1/u = 1/(-15) + 1/(-10) = -25/150 = -1/6
- v = -6 cm: negative, so the image is on the object side and virtual.
- m = v/u = (-6)/(-10) = +0.6: positive, so erect, and diminished.
- h′ = m × h = +0.6 × +2 = +1.2 cm
- Power P = 1/f = 1/(-0.15 m) = -6.67 D
Where the formula comes from
- Take an object AB in front of a convex lens and draw a ray from A parallel to the axis, which refracts through F₂, and one through the optical centre O, which goes straight. They meet at A′.
- Triangles A′B′O and ABO are similar, so A′B′/AB = OB′/OB.
- The parallel ray meets the lens at a height equal to AB, so triangles A′B′F₂ and the one at the lens give A′B′/AB = B′F₂/OF₂ = (OB′ - OF₂)/OF₂.
- Equate: OB′/OB = (OB′ - OF₂)/OF₂.
- With signs, OB = -u, OB′ = v, OF₂ = f. So v/(-u) = (v - f)/f, which gives uv = uf - vf.
- Divide by uvf: 1/v - 1/u = 1/f.
New Cartesian signs
- The object is always placed to the left of the mirror or lens, so light falls on it from the left.
- All distances parallel to the principal axis are measured from the pole of the mirror or the optical centre of the lens.
- Distances measured to the right of the origin, along the direction of the incident light, are positive (+x).
- Distances measured to the left of the origin, against the incident light, are negative (-x).
- Distances measured upward and perpendicular to the principal axis, such as the height of an erect object or image, are positive (+y).
- Distances measured downward, such as the height of an inverted image, are negative (-y).
Quick answers
What is the lens formula?
1/v - 1/u = 1/f, relating the object distance u, the image distance v and the focal length f, all measured from the optical centre O.
Why is there a minus sign in the lens formula but a plus in the mirror formula?
A lens sends light on to the far side, so a real image has a positive v. A mirror sends it back, so a real image has a negative v. The signs of the convention change the form of the equation.
What is the image for a concave lens of focal length 15 cm with the object at 10 cm?
v = -6 cm and m = +0.6: virtual, erect and diminished, on the same side as the object.