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Geometric Optics · Image Formation by Mirrors and Lenses

Refraction and Thin Convex Lenses

Why can Jayden's magnifying glass make a candle flame show up on a sheet of paper, upside down, when the flame is far away, but make a coin look bigger and upright when it is close? The answer is in how a convex lens bends light. By the end of this page you will be able to predict where an image forms and what it looks like.

Refraction and the Convex Lens

When light passes from air into glass, it slows down and changes direction at the surface. This bending is called refraction. How much a material slows light is measured by its index of refraction $n=c/v$, and the bending follows Snell's law: $n_1\sin\theta_1=n_2\sin\theta_2$. Going from air into glass, the ray bends toward the normal. A thin convex lens (also called a converging lens) is thicker in the middle than at the edges. Each ray bends toward the thick part, so rays that enter side by side are pulled together.

A beam of rays parallel to the principal axis (the line through the center of the lens, perpendicular to it) meets at one point after passing through the lens.

Parallel rays pass through a convex lens and meet at the focal point on the other side. F F f principal axis

Why is it called "converging"? Because the rays come together. A lens with a shorter focal length bends light more strongly.

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Define the focal point and focal length of a thin convex lens. What happens to a beam of rays parallel to the principal axis after it passes through the lens?

Define the focal point and focal length of a thin convex lens. What happens to a beam of rays parallel to the principal axis after it passes through the lens?

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Principal Rays and Image Types

To find an image, you only need to draw two or three special rays from the top of the object. Where they meet is where the top of the image forms.

  1. Parallel ray: travels parallel to the principal axis, then refracts through the far focal point.

  2. Central ray: goes through the center of the lens and keeps going straight (a thin lens does not bend it).

  3. Focal ray: passes through the near focal point, then refracts to travel parallel to the axis.

Ray diagram for an object beyond twice the focal length: the image is real, inverted, and smaller. F F 2F 2F object image

The image can be described with two pairs of words:

  • Real image: light rays actually meet there. You can project it on a card or screen.

  • Virtual image: rays only appear to come from there. It cannot be projected, and you see it by looking through the lens.

  • Upright (same orientation as the object) or inverted (upside down).

The type of image depends on where the object is compared with $f$ and $2f$:

Object position Image
Beyond $2f$ Real, inverted, reduced
At $2f$ Real, inverted, same size
Between $f$ and $2f$ Real, inverted, enlarged
At $f$ No image (rays leave parallel)
Between lens and $f$ Virtual, upright, enlarged
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Classify each image formed by a convex lens as real or virtual and as upright or inverted: (a) an object placed beyond $2f$, and (b) an object placed between the lens…

Classify each image formed by a convex lens as real or virtual and as upright or inverted: (a) an object placed beyond $2f$, and (b) an object placed between the lens and the focal point $f$.

Solve

The Thin Lens Equation and Magnification

Drawings give the idea. Numbers give the exact answer. For a thin lens:

The signs tell you what kind of image you have:

  • $f>0$ for a convex lens, and $d_o>0$ for a real object.

  • $d_i>0$: real image (on the opposite side from the object).

  • $di<0$: virtual image (on the same side as the object).

  • $m<0$: inverted. $m>0$: upright. $\lvert m\rvert>1$ means enlarged; $\lvert m\rvert<1$ means reduced.

Why the minus sign in $m$? A real image ($d_i>0$) is always inverted, so a negative $m$ records that flip.

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A thin convex lens has a focal length of $0.20\ \text{m}$. An object is placed $0.60\ \text{m}$ from the lens. Use the thin lens equation…

A thin convex lens has a focal length of $0.20\ \text{m}$. An object is placed $0.60\ \text{m}$ from the lens. Use the thin lens equation $\dfrac{1}{f}=\dfrac{1}{d_o}+\dfrac{1}{d_i}$ to find the image distance.

Solve
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Jayden uses a convex lens with a focal length of $15\ \text{cm}$ to project an image of a candle flame onto a card. The candle is $45\ \text{cm}$ from the lens. (a) Find…

Jayden uses a convex lens with a focal length of $15\ \text{cm}$ to project an image of a candle flame onto a card. The candle is $45\ \text{cm}$ from the lens. (a) Find the image distance. (b) Find the magnification $m=-\dfrac{d_i}{d_o}$. (c) If the flame is $2.0\ \text{cm}$ tall, find the height of its image and state whether it is upright or inverted.

Solve
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Emma uses a convex lens with $f=10\ \text{cm}$ as a magnifying glass. She holds a coin $6.0\ \text{cm}$ from the lens. (a) Find the image distance, including its sign.…

Emma uses a convex lens with $f=10\ \text{cm}$ as a magnifying glass. She holds a coin $6.0\ \text{cm}$ from the lens. (a) Find the image distance, including its sign. (b) State whether the image is real or virtual and upright or inverted. (c) Find the magnification.

Solve

Putting It Together: Checking a Drawing and Moving the Object

A good habit: use the ray diagram first to predict the type of image, then use the equation to get numbers and check the drawing.

Now watch what happens as the object slides toward the lens. Far away, the image is small, real, and close to $F$. As the object approaches $2f$, the image grows and moves out. At $f$ the image runs off to infinity. Past that, the image jumps to the same side as the object and becomes virtual, upright, and enlarged.

Ray diagram for an object inside the focal length: the rays diverge and the image is virtual, upright, and enlarged. F F object virtual image

The dashed lines show where the rays seem to come from. Your eye traces them back and sees a larger, upright image behind the object.

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A convex lens has a focal length of $12\ \text{cm}$. An object $3.0\ \text{cm}$ tall stands $20\ \text{cm}$ from the lens. Describe how to draw the three principal rays…

A convex lens has a focal length of $12\ \text{cm}$. An object $3.0\ \text{cm}$ tall stands $20\ \text{cm}$ from the lens. Describe how to draw the three principal rays (parallel ray, central ray, focal ray) to locate the image. Then use the thin lens equation to check the image distance and find the image height.

Solve
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A projector uses a thin convex lens with $f=0.10\ \text{m}$ to put an image of a slide $0.036\ \text{m}$ tall onto a screen. The screen is $4.0\ \text{m}$ from the lens.…

A projector uses a thin convex lens with $f=0.10\ \text{m}$ to put an image of a slide $0.036\ \text{m}$ tall onto a screen. The screen is $4.0\ \text{m}$ from the lens. (a) Find how far the slide must be from the lens. (b) Find the height of the image on the screen and its orientation relative to the slide. (c) Explain why the slide is loaded upside down to make the picture appear upright. (d) The projector is moved so the screen is $8.0\ \text{m}$ away. Must the lens move toward or away from the slide to refocus? Justify your answer.

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Mateo places an object on the principal axis of a thin convex lens with $f=8.0\ \text{cm}$ and studies how the image changes as he moves the object. For the object…

Mateo places an object on the principal axis of a thin convex lens with $f=8.0\ \text{cm}$ and studies how the image changes as he moves the object. For the object distances $d_o=24\ \text{cm}$, $16\ \text{cm}$, $12\ \text{cm}$, $8.0\ \text{cm}$, and $5.0\ \text{cm}$, determine the image distance (or state that no image forms at a finite distance), and classify each image as real or virtual, upright or inverted, enlarged or reduced. Then explain, using ray diagrams, why the image changes from real to virtual as the object crosses the focal point.

Solve

Back to the start: Jayden's candle was farther from the lens than $f$ (at $3f$), so the image was real, inverted, and could be caught on a card. Emma's coin was inside $f$, so the image was virtual, upright, and larger.

Summary

See all the exercises in this section