Thin lenses and the lens maker's equation
Use thin-lens and lensmaker equations in the small-angle approximation. Apply explicit distance and curvature signs to find focal length, finite image position and magnification, and distinguish real from virtual images.
What a learner can do afterwards
- Uses the lens equation and a consistent sign convention to locate an image and find its magnification
- Relates focal length to surface curvature and refractive index
- Distinguishes real from virtual images and says which can be caught on a screen
1 · Read
You have traced rays through a converging lens. Use an ideal thin lens in air, with rays close to the optical axis and making small angles with it. A central ray continues straight. For a converging lens, an axis-parallel ray leaves through the far focus. A ray whose incoming line passes through the near focus leaves parallel to the axis.
For a real object before the lens, object distance do is positive. Focal length f is positive for a converging lens and negative for a diverging lens. Image distance di is positive for a real image on the outgoing side and negative for a virtual image on the object's side. Measure distances from the lens centre. Use 1/f = 1/do + 1/di, so 1/di = 1/f - 1/do. Keep length units consistent.
Calculate reciprocal image distance first, then invert it. For f = +8 cm and do = +24 cm, 1/di = 1/8 - 1/24 = 2/24 = 1/12 per cm. Thus di = +12 cm, a real image on the outgoing side. Keep any negative sign when taking a reciprocal.
These cells show a calculation, not a ray diagram.
Magnification m = -di/do is image height divided by object height. Positive m means upright; negative m means inverted. Its magnitude gives the size ratio: |m| > 1 is enlarged, |m| = 1 is same-size, and |m| < 1 is reduced. For do = +25 cm and di = +10 cm, m = -10/25 = -0.4: inverted and 0.4 times the object's height.
Trace rays from the same object point. Where the actual outgoing rays cross, they form a real image that a screen can receive at its image plane. If only backward extensions meet, they locate a virtual image. That lens alone cannot project it onto a screen at its apparent position. An empty screen does not prove an image is virtual: the screen might be misplaced. Another optical system can form a new image from the outgoing light.
For a converging lens with 0 < do < f, the object is inside the focal length. Then 1/di = 1/f - 1/do is negative, so the image is virtual. Also m = -di/do = f/(f-do), which is positive and greater than one: upright and enlarged. At do = f, rays from each object point leave parallel and no finite image forms.
For a thin spherical lens, 1/f = (nL/nM - 1)*(1/R1 - 1/R2). nL is lens index; nM is the same surrounding index on both sides. Let light travel left to right. R1 belongs to the first surface and R2 to the second. A radius is positive if its curvature centre is right of that surface and negative if left. A flat face contributes zero reciprocal curvature.
In air, take nM = 1 and write n = nL. Then 1/f = (n-1)*(1/R1 - 1/R2). For n = 1.6, R1 = +30 cm and R2 = -30 cm, 1/f = 0.6*(1/30 + 1/30) = 0.04 per cm. Thus f = +25 cm. Subtract the signed second reciprocal without losing its minus sign.
These are arithmetic steps, not a drawing of the lens surfaces.
For a fixed-f single-lens camera model with f > 0 and do > f, di = f*do/(do-f) = f + f*f/(do-f). Moving the object closer reduces do-f, so di increases. Refocus by increasing the lens-to-sensor separation. For a very distant object, di approaches f. Real cameras may use several lens elements and other focusing arrangements.
Keep distance and curvature signs. The lens equation locates finite images; magnification gives orientation and size. Actual outgoing ray crossings form real images.
Lesson, questions and flashcards written for this dataset.
Readings and flashcards written for this dataset. Includes links to Khan Academy's "Convex lenses" and "Lens formula". Video licenses are not specified.
More to read
Keep the sign when taking a reciprocal
For the distance convention used here, a real object has do > 0. A converging lens has f > 0. A real outgoing-side image has di > 0; a virtual image on the object's side has di < 0.
Take f = +6 cm and do = +18 cm. Then 1/di = 1/6 - 1/18 = 2/18 per cm, so di = +9 cm. Magnification is -9/18 = -0.5. The image is real, inverted and half-size.
Now put a real object inside a converging lens's focal length. The reciprocal difference becomes negative. Its reciprocal must remain negative too. Magnification is positive, and f/(f-do) shows that its magnitude exceeds one.
Some resources use signed coordinates u and v instead. If a resource defines u = -do and v = di, its equation 1/v - 1/u = 1/f becomes the equation used here. Check the resource's definitions before translating. The supplied record for the linked "Lens formula" video uses that equation, but its full definitions still need checking.
Two surfaces determine the focal length
A spherical refracting surface has a different local normal at each point. A lens refracts light at entry and exit. For a thin lens and rays making small angles with the optical axis, the combined effect is described by the lensmaker equation.
Use 1/f = (nL/nM - 1)*(1/R1 - 1/R2), where nL is the lens index and nM is the common surrounding index on both sides. Let light travel left to right. A surface radius is positive if its curvature centre lies to the right of that surface and negative if it lies to the left. Number the surfaces in the order light meets them.
A flat face contributes zero reciprocal curvature. Equal lens and surrounding indices make the index factor zero, giving no refracting power in this model. Changing the surrounding material can therefore change focal length without changing the lens shape.
Focal length does not specify every optical property. Aperture, transmitted brightness and image defects need information beyond the thin-lens distance equation.
Refocusing a one-lens camera model
Model a camera with one ideal converging lens and a sensor. Hold positive focal length f fixed and keep the object beyond f. The equation gives di = f + f*f/(do-f). The sensor must lie at that real image distance for a sharp image in this model.
As the object approaches the focus from beyond it, do-f decreases while remaining positive. The fraction increases, so the image plane moves farther from the lens. Refocus by increasing lens-to-sensor separation. For an object very far away, the fraction becomes small and di approaches f.
This does not describe every camera's focusing mechanism. Real camera optics may contain several elements, with only some elements moving during focusing. A virtual image also does not mean that no camera can record the outgoing light: another optical system can use that light to form a new real image on its sensor.
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