4.4 Thin Lens Equation and Magnification

Thin Lens Equation Derivation

Lensmaker's Equation and Thin Lens Approximation

The thin lens equation comes from a more general result called the lensmaker's equation, which relates a lens's focal length to its index of refraction and the radii of curvature of its two surfaces. To get from the lensmaker's equation to the thin lens equation, you make two simplifying assumptions:

1. The lens thickness is negligible compared to the focal length and radii of curvature.
2. Incoming rays make small angles with the optical axis (the small-angle approximation).

With those assumptions, you can trace light refracting at each surface separately, then combine the results. What falls out is the thin lens equation:

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

where $f$ is the focal length, $d_o$ is the object distance, and $d_i$ is the image distance. This single equation works for both converging and diverging lenses as long as you follow the sign conventions (covered below).

The derivation is worth understanding, not just memorizing. It shows how the curvature of each surface contributes to the overall bending of light, and why focal length depends on both the lens shape and the material's refractive index.

Applications and Limitations

The thin lens equation is the starting point for analyzing simple optical systems like cameras, eyeglasses, and magnifying glasses. It gives accurate results whenever the lens is thin relative to the other distances involved.

Where it breaks down:

• Thick lenses where you can't ignore the separation between the two refracting surfaces
• Multi-element systems (like in microscope objectives or camera zoom lenses) where interactions between elements matter
• Wide-angle rays that violate the small-angle approximation, leading to aberrations

Knowing the derivation helps you recognize when the approximation is trustworthy and when you need a more sophisticated model.

Solving Thin Lens Problems

Equation Manipulation and Sign Conventions

You can rearrange $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ to isolate whichever quantity you need. For example, solving for image distance:

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

The sign conventions are critical and trip people up constantly. Here's the standard set:

After you solve for $d_i$, its sign tells you the image type: positive means real, negative means virtual.

A good habit is to pair your calculation with a ray diagram. Draw at least two principal rays (parallel ray, focal ray, or central ray) to confirm that your math and your picture agree.

Problem-Solving Strategies and Applications

For multi-lens systems, apply the thin lens equation one lens at a time. The image formed by the first lens becomes the object for the second lens. If that "object" falls behind the second lens, treat it as a virtual object with a negative $d_o$.

Common problem types to practice:

• Converging lens, distant object: Object far away ($d_o \gg f$), so the image forms near the focal point on the far side. This is how a camera sensor captures a scene.
• Diverging lens, nearby object: Image is always virtual, upright, and reduced. This is the principle behind corrective lenses for nearsightedness.
• Two-lens system: A compound microscope uses an objective lens to form a real, enlarged intermediate image, which the eyepiece then magnifies further.

For vision correction problems, remember that lens power in diopters is $P = \frac{1}{f}$ where $f$ is in meters.

Linear vs Angular Magnification

Linear Magnification

Linear (lateral) magnification describes how the image size compares to the object size. It's defined two equivalent ways:

$M = \frac{h_i}{h_o} = -\frac{d_i}{d_o}$

The negative sign in $-d_i/d_o$ accounts for image inversion. Interpreting the result:

• $M = -2$: the image is inverted and twice as tall as the object
• $M = +0.5$: the image is upright and half the height of the object
• The sign tells you orientation (positive = upright, negative = inverted), and the magnitude tells you the size ratio.

Angular Magnification

Angular magnification compares the angle an image subtends at your eye to the angle the object would subtend when viewed without the instrument (typically from the near point, about 25 cm). For a simple magnifying glass:

$m = \frac{\theta_{\text{image}}}{\theta_{\text{unaided}}}$

This is the quantity that matters for instruments you look through, because your eye perceives size based on angular extent, not physical height. A telescope with $m = 50\times$ makes a distant star's angular separation appear 50 times larger. A magnifying glass rated at $10\times$ makes fine print subtend an angle 10 times bigger than it would at your near point.

Relationship and Applications

Linear and angular magnification measure different things, and which one matters depends on the instrument:

• Microscopes use both. The objective lens produces a linearly magnified real image, and the eyepiece provides additional angular magnification. Total magnification is the product of the two.
• Telescopes are characterized by angular magnification because the objects are so far away that linear image size alone doesn't capture how much "closer" they appear.
• Projectors and cameras care about linear magnification since they produce a physical image on a screen or sensor.

Image Magnification Calculation

Magnification Formulas for Lenses and Mirrors

Both lenses and mirrors use the same magnification formula:

$M = \frac{h_i}{h_o} = -\frac{d_i}{d_o}$

Quick reference for interpreting $M$:

For mirrors, the same sign conventions apply: positive $d_i$ for real images (formed in front of the mirror), negative for virtual images (behind the mirror).

Complex Systems and Practical Applications

For a system with multiple optical elements, the total magnification is the product of each element's individual magnification:

$M_{\text{total}} = M_1 \times M_2 \times \cdots \times M_n$

A compound microscope with a $10\times$ objective and a $20\times$ eyepiece gives $M_{\text{total}} = 200\times$.

Worked example: A converging lens with $f = 15 \text{ cm}$ has an object placed at $d_o = 25 \text{ cm}$. Find the image distance and magnification.

1. Apply the thin lens equation: $\frac{1}{d_i} = \frac{1}{15} - \frac{1}{25} = \frac{5 - 3}{75} = \frac{2}{75}$

2. Solve: $d_i = 37.5 \text{ cm}$ (positive, so the image is real)

3. Find magnification: $M = -\frac{37.5}{25} = -1.5$

4. Interpret: The image is real, inverted, and 1.5 times the height of the object.

This same approach scales to any problem, whether you're designing a projector, sizing a camera sensor, or figuring out the prescription for a pair of glasses.