Physics · Unit 6: Optics ·  Activity 6-1-3

Deep Dive: Lenses — Converging & Diverging

🔬 Deep Dive
This is your textbook for this topic. Read it alongside the activity.
AConcept

What a Lens Does

A lens is a piece of clear material with at least one curved surface. Light refracts when it enters and again when it leaves, and the curve makes rays that hit different parts of the lens bend by different amounts. Think of a lens as a stack of tiny prisms: the ones near the edge bend light a lot, and the ones near the middle hardly bend it at all.

A converging (convex) lens is thicker in the middle. Its prisms point so that every ray bends toward the axis, and rays that arrive parallel to the axis all meet at the focal point, a distance f(the focal length) from the lens. A diverging (concave) lens is thinner in the middle. It bends every ray away from the axis, so parallel rays spread out as if they came from a focal point on the incoming side.

A lens has a focal point on each side, at the same distance f. We give f a sign: positive for converging and negative for diverging.

BConcept

Where the Image Forms

Two rays are enough to find an image. A ray parallel to the axis bends through the far focal point, and a ray through the center of the lens goes straight on. Where the rays meet is the tip of the image. For a converging lens the answer depends on where the object is, exactly as with a concave mirror:

Object positionImageExample
d_o > 2fReal, inverted, smallerA camera, an eye
d_o = 2fReal, inverted, the same size
f < d_o < 2fReal, inverted, largerA projector, a slide viewer
d_o = fNo image: the rays emerge parallelA lighthouse or spotlight lens
d_o < fVirtual, upright, largerA magnifying glass

A real image is where light actually meets, on the far side of the lens. A screen placed there shows it. A virtual image is where light only seems to come from, so no screen can catch it. A diverging lens works the same way at every distance: its image is always virtual, upright, smaller, and between the lens and the object.

CMath

The Thin-Lens Equation

It is the same pair of equations you met for curved mirrors:

1/f = 1/d_o + 1/d_i
m = h_i ÷ h_o = −d_i ÷ d_o

The signs again carry the meaning. f is positive for converging and negative for diverging. d_i is positive for a real image (on the far side of the lens) and negative for a virtual one (on the object’s side). The magnification mis negative for an inverted image, and |m| > 1 means the image is larger.

Example 1: a converging lens, object far away
Given
  • f = +10 cm, d_o = 30 cm, h_o = 4.0 cm
Solve
1/d_i = 1/f − 1/d_o = 1/10 − 1/30 = 2/30
d_i = +15 cm
m = −d_i ÷ d_o = −15 ÷ 30 = −0.50
h_i = m × h_o = −2.0 cm
✓ A real image 15 cm beyond the lens, inverted and half the size.
Example 2: a converging lens as a magnifier
Given
  • f = +10 cm, d_o = 5.0 cm, h_o = 4.0 cm
Solve
1/d_i = 1/10 − 1/5 = −1/10
d_i = −10 cm
m = −(−10) ÷ 5.0 = +2.0
h_i = +8.0 cm
✓ A virtual image 10 cm on the object’s side, upright and twice the size.
Example 3: a diverging lens
Given
  • f = −20 cm, d_o = 40 cm, h_o = 4.0 cm
Solve
1/d_i = 1/f − 1/d_o = −1/20 − 1/40 = −3/40
d_i = −13.3 cm
m = −(−13.3) ÷ 40 = +0.33
h_i = +1.3 cm
✓ A virtual image 13.3 cm on the object’s side, upright and a third of the size.
DExample

Lens Explorer

Start with a converging lens and f = 10 cm. Slide the object in from far away and watch the real image grow and move away, then jump to the object’s side as a virtual, magnified image when you cross f. Then switch to a diverging lens.

Pick a lens, set its focal length, and slide the object toward and away from it. Two rays leave the tip of the object: one parallel to the axis, and one through the center of the lens, which goes straight on.

FFobjectimage
Lens:
f = +10 cm  ·  1/f = 1/d_o + 1/d_i
d_i = 15.0 cm  ·  m = −d_i/d_o = -0.50
The image is real, inverted, and smaller.
EMath

Lens Power and Diopters

Opticians don’t describe a lens by its focal length. They use its power, in diopters (D):

P = 1 ÷ f (f in meters)

A shorter focal length is a stronger lens, so it has a bigger power. A converging lens has a positive power and a diverging lens a negative one. And powers of thin lenses held together simply add.

Reading a prescription
Given
  • A converging lens has f = +0.50 m
  • A diverging lens is labeled −4.0 D
Solve
P = 1 ÷ 0.50 = +2.0 D
f = 1 ÷ (−4.0) = −0.25 m = −25 cm
Both lenses together: +2.0 + (−4.0) = −2.0 D
✓ The first is a +2.0 D converging lens. The second has a focal length of −25 cm, and the pair together act as a −2.0 D diverging lens.
FConcept

The Eye and Glasses

Your eye is a converging lens system. The cornea and the lens inside the eye bend light so that the world forms a real, inverted image on the retina, the light-sensing layer at the back. Your brain flips it back over. For a sharp picture, the image has to land on the retina.

In a nearsighted eye the focus lands in front of the retina, so distant things are blurry. The eye is effectively too strong for its length. A diverging glasses lens spreads the light a little before it enters the eye and pushes the focus back onto the retina. In a farsighted eye the focus would land behind the retina, so close things are blurry. A converging glasses lens pulls the focus forward.

🔑That is the unit’s second question. Glasses don’t sharpen the picture themselves. They shift where the eye’s own lens focuses, until the focus is on the retina again.
Correcting a nearsighted eye (a simplified model)
Given
  • Eye: lens-to-retina distance 17.0 mm, and the eye’s own power is 62.0 D
  • Glasses lens touching the eye, with powers adding
Solve
Focus without glasses: f = 1000 ÷ 62.0 = 16.1 mm, which is 0.9 mm in front of the retina
The retina needs a total power of 1000 ÷ 17.0 = 58.8 D
Glasses power = 58.8 − 62.0 = −3.2 D (about −3.25 D)
Check: 62.0 + (−3.25) = 58.75 D, so f = 1000 ÷ 58.75 = 17.0 mm ✓
✓ A −3.25 D diverging lens puts the focus exactly on the retina.
GExample

Eye and Glasses Explorer

Pick an eye type, look at where the focus lands, and then choose a glasses lens that makes the picture sharp. Try the nearsighted eye with a converging lens, and see what goes wrong.

Light from a distant object enters the eye as parallel rays. The eye’s lens (with the glasses, if any) focuses them, and the picture is sharp only if the focus lands on the retina. Start with a nearsighted eye, then add a glasses lens until the picture is sharp. (The vertical scale is exaggerated.)

eye lensretinafocus
Eye type:
Total power = 62.0 + (0.00) = 62.00 D
Focus distance = 1000 ÷ 62.00 = 16.13 mm  ·  retina at 17.0 mm
The focus is 0.87 mm in front of the retina
Blurry. Each point of light is smeared into a spot about 0.16 mm across. The eye is too strong, so a diverging (negative) lens would push the focus back.
💡This is a simplified model. A real eye has several surfaces and the lens inside can change shape to focus on near objects. But the core idea is the same, and it’s the one you’ll need in Problem 6-2-1.
HMath

Lenses Working Together

Project 6-1-4 can use lenses in pairs. Two simple results cover most of what you’ll build:

A magnifier. Hold the object at the focal point, and the rays leave the lens parallel, so your relaxed eye sees a large virtual image. How much bigger it looks (its angular magnification) is about

M ≈ 25 cm ÷ f

where 25 cm is the usual closest comfortable viewing distance. A lens with f = 5 cm magnifies about 5×.

A simple telescope. The objective, a long-focal-length converging lens, makes a real image of the distant object at its focal point. The eyepiece, a short-focal-length lens, acts as a magnifier on that image. When the lenses are one total focal length apart,

M = f_objective ÷ f_eyepiece length ≈ f_objective + f_eyepiece
Planning a telescope
Given
  • Objective f = 30 cm, eyepiece f = 5.0 cm
Solve
M = 30 ÷ 5.0 = 6×
Tube length ≈ 30 + 5.0 = 35 cm
✓ A 6× telescope about 35 cm long. The view is inverted, which doesn’t matter for stars but does for birdwatching.
IWatch Out

Three Common Mix-Ups

⚠️“A converging lens always makes a real, inverted image.” Only if the object is farther away than the focal length. Inside f, a converging lens makes a virtual, upright, magnified image, and that’s exactly what a magnifying glass does.
⚠️“A diverging lens can focus an image on a screen.” It can’t. A diverging lens only makes virtual images, from light that spreads out instead of meeting. There is nowhere for a screen to catch it.
⚠️“Cover half the lens and half the image disappears.” It doesn’t. Every part of the lens sends light from every point of the object to the matching image point. Covering half the lens just makes the whole image dimmer.
JConcept

Quick Check

Four questions. They are not graded. Answer them to find out what still needs work.

Q1A converging lens has f = +12 cm, and an object is 36 cm away. Where does the image form?
Q2Which lens corrects nearsightedness?
Q3What is the focal length of a lens with a power of +4.0 D?
Q4An object is placed inside the focal length of a converging lens. What is the image like?
Score: 0 of 4
KConcept

Looking Ahead

You now have all three of light’s tricks: bounce, bend, and focus. Next you’ll put them to work.

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