Lenses

Converging and diverging lenses, thin lens equation, magnification, optical instruments

🔍 Lenses

Types of Lenses

Converging (Convex) Lens:

Thicker in middle
Focal length f > 0 (positive)
Parallel rays converge at focal point
Can form real or virtual images

Diverging (Concave) Lens:

Thinner in middle
Focal length f < 0 (negative)
Parallel rays appear to diverge from focal point
Always forms virtual, upright, reduced images

Both sides of lens have same focal length!

Thin Lens Equation

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

where:

$f$ = focal length (+ for converging, - for diverging)
$d_o$ = object distance (always positive)
$d_i$ = image distance (+ for real, - for virtual)

Magnification: $m = -\frac{d_i}{d_o} = \frac{h_i}{h_o}$

💡 Same equations as mirrors! But sign conventions differ slightly.

Sign Conventions (Lenses)

| Quantity | Positive | Negative | |----------|----------|----------| | $f$ | Converging | Diverging | | $d_o$ | Real object | (rare) | | $d_i$ | Real (opposite side) | Virtual (same side as object) | | $m$ | Upright | Inverted |

Key difference from mirrors: Real image on opposite side of lens from object!

Ray Diagrams (Converging Lens)

Draw any 2 of these 3 rays:

Parallel ray → refracts through F on far side
Focal ray (through F on near side) → refracts parallel
Center ray (through lens center) → straight through (no bend)

Where rays intersect = image location!

Cases:

$d_o > 2f$ : Real, inverted, reduced (cameras)
$d_o = 2f$ : Real, inverted, same size
$f < d_o < 2f$ : Real, inverted, enlarged (projectors)
$d_o < f$ : Virtual, upright, enlarged (magnifying glass!)

Lensmaker's Equation

$\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

where:

n = index of refraction of lens
$R_1$ , $R_2$ = radii of curvature of surfaces

(Not commonly used in AP Physics 2, but good to know!)

Power of Lens

$P = \frac{1}{f}$

Unit: Diopter (D) = m⁻¹

Higher power → shorter f → more bending

Example: Eyeglass prescription "+2.00 D" means f = 0.50 m

Compound Lens Systems

Two lenses in combination:

Method 1 (Total power): $P_{total} = P_1 + P_2$

Method 2 (Step-by-step):

Find image from lens 1
Use that image as object for lens 2
Total magnification: $m_{total} = m_1 \times m_2$

Optical Instruments

Magnifying Glass:

Single converging lens
Object inside focal length ( $d_o < f$ )
Virtual, upright, enlarged image
Angular magnification: $M = \frac{25 \text{ cm}}{f}$

Compound Microscope:

Objective lens (short f): Forms real, enlarged image
Eyepiece lens: Acts as magnifying glass on that image
Total magnification: $M = m_o \times M_e = -\frac{L}{f_o} \times \frac{25}{f_e}$
Very high magnification (~100-1000×)

Telescope:

Objective lens (long f): Forms real, reduced image of distant object
Eyepiece: Magnifies that image
Angular magnification: $M = -\frac{f_o}{f_e}$
Larger objective → more light → see fainter objects

Camera:

Converging lens
Object far away ( $d_o >> f$ ), so $d_i \approx f$
Real, inverted, reduced image on film/sensor
Adjustable f (zoom) and aperture (brightness)

Human Eye:

Cornea + lens = converging system
Lens changes shape (accommodation) to focus
Image on retina (real, inverted, reduced)
Brain flips it!

Nearsighted (myopia): Eyeball too long → use diverging lens Farsighted (hyperopia): Eyeball too short → use converging lens

Aberrations

Spherical aberration: Rays at edge focus differently

Solution: Parabolic lens, smaller aperture

Chromatic aberration: Different colors focus at different points

Solution: Achromatic doublet (two lenses)

Problem-Solving Strategy

Identify lens type: f > 0 (converging) or f < 0 (diverging)
Use thin lens equation: $1/f = 1/d_o + 1/d_i$
Find magnification: $m = -d_i/d_o$
Interpret signs:
- $d_i > 0$ : Real (opposite side)
- $d_i < 0$ : Virtual (same side)
- $m < 0$ : Inverted
- $m > 0$ : Upright

Common Mistakes

❌ Confusing lens and mirror sign conventions (real image location!) ❌ Forgetting f is negative for diverging lens ❌ Using wrong magnification formula for instruments ❌ Thinking diverging lens can form real image (never!) ❌ Not checking calculator mode (degrees vs radians)

📚 Practice Problems

1Problem 1easy

❓ Question:

A converging lens with focal length 15 cm is used to form an image of an object 30 cm away. Find (a) image distance, (b) magnification, (c) describe image.

💡 Show Solution

Given:

Focal length: $f = 15$ cm (positive, converging)
Object distance: $d_o = 30$ cm

Part (a): Image distance

Thin lens equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $\frac{1}{15} = \frac{1}{30} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{15} - \frac{1}{30} = \frac{2-1}{30} = \frac{1}{30}$ $d_i = 30 \text{ cm}$

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{30} = -1.0$

Part (c): Image description

$d_i > 0$ : Real (opposite side of lens)
$m < 0$ : Inverted
$|m| = 1$ : Same size as object
Note: Object is at 2f, image also at 2f!

Answer:

(a) d_i = 30 cm (opposite side)
(b) m = -1.0
(c) Real, inverted, same size

2Problem 2easy

❓ Question:

A converging lens with focal length 15 cm is used to form an image of an object 30 cm away. Find (a) image distance, (b) magnification, (c) describe image.

💡 Show Solution

Given:

Focal length: $f = 15$ cm (positive, converging)
Object distance: $d_o = 30$ cm

Part (a): Image distance

Part (b): Magnification

$m = -\frac{d_i}{d_o} = -\frac{30}{30} = -1.0$

Part (c): Image description

$d_i > 0$ : Real (opposite side of lens)
$m < 0$ : Inverted
$|m| = 1$ : Same size as object
Note: Object is at 2f, image also at 2f!

Answer:

(a) d_i = 30 cm (opposite side)
(b) m = -1.0
(c) Real, inverted, same size

3Problem 3medium

❓ Question:

A diverging lens has focal length -20 cm. An object is placed 40 cm from the lens. Find the image distance and magnification.

💡 Show Solution

Given:

Focal length: $f = -20$ cm (negative, diverging)
Object distance: $d_o = 40$ cm

Solution:

Thin lens equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $\frac{1}{-20} = \frac{1}{40} + \frac{1}{d_i}$ $\frac{1}{d_i} = -\frac{1}{20} - \frac{1}{40} = \frac{-2-1}{40} = -\frac{3}{40}$ $d_i = -\frac{40}{3} = -13.3 \text{ cm}$

Magnification: $m = -\frac{d_i}{d_o} = -\frac{(-13.3)}{40} = +0.33$

Image description:

$d_i < 0$ : Virtual (same side as object)
$m > 0$ : Upright
$|m| < 1$ : Reduced (1/3 size)

Answer:

d_i = -13.3 cm (same side, virtual)
m = +0.33 (upright, reduced)

Note: Diverging lenses ALWAYS produce virtual, upright, reduced images!

4Problem 4medium

❓ Question:

A diverging lens has focal length -20 cm. An object is placed 40 cm from the lens. Find the image distance and magnification.

💡 Show Solution

Given:

Focal length: $f = -20$ cm (negative, diverging)
Object distance: $d_o = 40$ cm

Solution:

Magnification: $m = -\frac{d_i}{d_o} = -\frac{(-13.3)}{40} = +0.33$

Image description:

$d_i < 0$ : Virtual (same side as object)
$m > 0$ : Upright
$|m| < 1$ : Reduced (1/3 size)

Answer:

d_i = -13.3 cm (same side, virtual)
m = +0.33 (upright, reduced)

Note: Diverging lenses ALWAYS produce virtual, upright, reduced images!

5Problem 5medium

❓ Question:

A converging lens has focal length 20 cm. An object is placed 30 cm from the lens. (a) Where is the image located? (b) What is the magnification? (c) Is the image real or virtual, upright or inverted?

💡 Show Solution

Solution:

Given: f = +20 cm (converging), d_o = +30 cm

(a) Image location (Thin lens equation): 1/f = 1/d_o + 1/d_i 1/20 = 1/30 + 1/d_i 1/d_i = 1/20 - 1/30 = 3/60 - 2/60 = 1/60 d_i = 60 cm (on opposite side of lens from object)

(b) Magnification: m = -d_i/d_o = -60/30 = -2.0

d_i > 0 → Real image (light actually converges)
m < 0 → Inverted
|m| = 2.0 → Image is twice as large as object

6Problem 6hard

❓ Question:

A converging lens with f = 10 cm is used as a magnifying glass. An object is placed 6 cm from the lens. Find the magnification.

💡 Show Solution

Given:

Focal length: $f = 10$ cm
Object distance: $d_o = 6$ cm

Note: Object is inside focal length ( $d_o < f$ ) - magnifying glass configuration!

Solution:

Find image distance: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $\frac{1}{10} = \frac{1}{6} + \frac{1}{d_i}$ $\frac{1}{d_i} = \frac{1}{10} - \frac{1}{6} = \frac{3-5}{30} = -\frac{2}{30} = -\frac{1}{15}$ $d_i = -15 \text{ cm}$

Magnification: $m = -\frac{d_i}{d_o} = -\frac{(-15)}{6} = +2.5$

Image description:

$d_i < 0$ : Virtual (same side as object)
$m > 0$ : Upright
$|m| > 1$ : Enlarged (2.5× larger!)

Answer: m = +2.5 (virtual, upright, enlarged)

This is why magnifying glasses work - object inside f gives enlarged virtual image!

7Problem 7hard

❓ Question:

A diverging lens has focal length -15 cm. An object is placed 10 cm from the lens. (a) Where is the image? (b) What is the magnification? (c) Describe the image.

💡 Show Solution

Solution:

Given: f = -15 cm (diverging), d_o = +10 cm

(a) Image location: 1/f = 1/d_o + 1/d_i 1/(-15) = 1/10 + 1/d_i 1/d_i = -1/15 - 1/10 = -2/30 - 3/30 = -5/30 = -1/6 d_i = -6.0 cm (same side as object)

(b) Magnification: m = -d_i/d_o = -(-6.0)/10 = +0.60

d_i < 0 → Virtual image (light appears to diverge from this point)
m > 0 → Upright
|m| = 0.60 → Reduced to 60% of object size

Diverging lenses always produce virtual, upright, reduced images.

8Problem 8hard

❓ Question:

A converging lens with f = 10 cm is used as a magnifying glass. An object is placed 6 cm from the lens. Find the magnification.

💡 Show Solution

Given:

Focal length: $f = 10$ cm
Object distance: $d_o = 6$ cm

Note: Object is inside focal length ( $d_o < f$ ) - magnifying glass configuration!

Solution:

Magnification: $m = -\frac{d_i}{d_o} = -\frac{(-15)}{6} = +2.5$

Image description:

$d_i < 0$ : Virtual (same side as object)
$m > 0$ : Upright
$|m| > 1$ : Enlarged (2.5× larger!)

Answer: m = +2.5 (virtual, upright, enlarged)

This is why magnifying glasses work - object inside f gives enlarged virtual image!

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