Images Formed by Lenses
Introduced| English | Meaning |
|---|---|
| lens/lenz/ | A transparent optical element that refracts light to form images or change a beam’s convergence. |
A magnifying glass, a camera, your own eye
- Hold a magnifying glass over a page and the letters swell; a camera shrinks a whole scene onto a sensor.
- Both use a lens 透镜, a shaped piece of glass that bends light by refraction.
- By bending rays to meet, a lens builds an image of whatever it looks at.
- Two simple rays are all you need to find where that image sits.
Converging and diverging lenses
- A converging (convex) lens is thicker in the middle; it bends parallel rays together to a focus.
- A diverging (concave) lens is thinner in the middle; it spreads parallel rays apart.
- The focal length $f$ is the distance from the lens to its focal point.
- A converging lens is the workhorse of cameras, projectors and the human eye.
A converging (convex) lens is:
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A converging lens is thicker in the middle and bends parallel rays to a focus.
Drawing the ray diagram
- Ray 1: from the top of the object, parallel to the axis, then bent through the far focus.
- Ray 2: from the top of the object, straight through the centre of the lens (undeviated).
- Where the two rays cross is the top of the image.
- Beyond the focal point the image is real and inverted; inside it, virtual and upright (a magnifier).
Build the lens image
Move the object nearer and farther than the focal point and watch the image switch from real to virtual.
An object beyond the focal length of a converging lens forms an image that is:
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Beyond $f$, the rays cross to form a real, inverted image.
Select all correct statements about lenses.
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Converging lenses focus rays (the diverging one spreads them); the centre ray is undeviated; inside f gives a magnifier.
The lens equation
- Object distance $u$, image distance $v$ and focal length $f$ obey $\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$.
- Solve it to find exactly where the image forms and how big it is.
- The magnification is the ratio of image to object height (or $v/u$).
- Camera and eye both move the lens or change $f$ to keep the image sharp.
An object is $6\ \text{cm}$ from a converging lens of $f = 2\ \text{cm}$. What is the image distance $v$, in cm?
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$1/v = 1/2 - 1/6 = 2/6 \Rightarrow v = 3\ \text{cm}$.
The lens equation is $\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{\_\_}$ (fill in the symbol).
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$1/f = 1/u + 1/v$ — object, image and focal distances.
A converging lens does not always give a real, inverted image. When the object is inside the focal length, the image is virtual, upright and magnified — that is exactly how a magnifying glass works. The object must be beyond $f$ to get a real, inverted image.
A converging lens used as a magnifying glass has the object inside the focal length, giving a virtual, upright, magnified image.
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Inside $f$, the image is virtual, upright and enlarged — the magnifier case.
An object is $6\ \text{cm}$ from a converging lens of focal length $f = 2\ \text{cm}$. Find the image distance $v$.
- $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{2} - \dfrac{1}{6} = \dfrac{3-1}{6} = \dfrac{2}{6}$.
- $v = 3\ \text{cm}$ — a real, inverted image on the far side.
A converging lens bends parallel rays to a focus and forms an image; use the parallel ray and the centre ray to find it. Beyond $f$ the image is real and inverted; inside $f$ it is virtual, upright and magnified. The lens equation $\dfrac1f = \dfrac1u + \dfrac1v$ locates it.
Write from an image distance of 6 cm
- An object is 12 cm from a converging lens of focal length 4 cm.
- Find the image distance.
An object is 12 cm from a converging lens of focal length 4 cm. What is the image distance, in cm?
One quarter minus one twelfth is one sixth.
What is the image distance?
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The lens equation gives v equal to 6 cm. Other options — 3 cm (the lesson image distance); 12 cm (the object distance used again).