13.4 Images Formed by Lenses

Lenses are optical devices that bend light to form images. This section explores how convex and concave lenses refract light differently, creating real or virtual images. Understanding these principles is crucial for analyzing optical systems in everyday life and technology.

The thin-lens equation and magnification formula are key tools for quantifying image formation. Ray diagrams provide a visual method to predict image characteristics, helping us grasp how lenses manipulate light to produce various image types and sizes.

Image formation by lenses

Convex Lens Refraction

Convex lenses bulge outward in the middle and cause light rays to converge. When parallel light rays strike a convex lens, they bend inward and meet at a single point on the opposite side called the focal point.

• These lenses are also known as converging lenses because they cause light to come together
• The distance from the lens to the focal point is called the focal length
• Convex lenses have positive focal lengths in optical calculations
• Common examples include magnifying glasses, camera lenses, and the lenses in telescopes

A magnifying glass demonstrates this principle when it focuses sunlight to a small, intense point that can become hot enough to start a fire. This happens because the lens concentrates the sun's parallel rays to a single focal point.

Concave Lens Refraction

Concave lenses curve inward in the middle and cause light rays to spread apart. When parallel light rays pass through a concave lens, they diverge (spread out) and appear to originate from a focal point on the same side that the light entered.

• These are called diverging lenses since they make light rays move away from each other
• The focal point of a concave lens is virtual—light doesn't actually pass through it
• Concave lenses have negative focal lengths in optical calculations
• They're commonly used in eyeglasses to correct nearsightedness by spreading out light before it reaches the eye

When you look through a concave lens at an object, it always appears smaller than it actually is, demonstrating the diverging effect on light rays.

Real Image Formation

Real images form when light rays actually converge at a location after passing through a lens. These images have distinctive properties that set them apart from virtual images.

• Light physically passes through the location where the image forms
• Real images can be projected onto a screen because light rays actually meet there
• They are typically inverted (upside-down) compared to the object
• Only convex lenses can form real images, and only when the object is beyond the focal point
• Examples include the image on a movie screen or the image formed on a camera's sensor

When a projector displays an image on a screen, you're seeing a real image formed by the projector's convex lens system.

Virtual Image Formation

Virtual images occur when light rays appear to diverge from a point but don't actually pass through that point. The brain interprets these diverging rays as coming from an apparent location.

• Light rays don't physically meet at the image location but appear to originate from there
• Cannot be captured on a screen since no light actually passes through the image location
• Usually appear upright (same orientation as the object)
• Both concave and convex lenses can form virtual images under certain conditions
• The image you see in a flat mirror is a common example of a virtual image

When you use a magnifying glass to examine something closely (with the object inside the focal length), you're looking at a virtual image that appears larger and is on the same side of the lens as the object.

Thin-Lens Equation

The thin-lens equation provides a mathematical relationship between the object distance, image distance, and focal length of a lens. It works for both convex and concave lenses.

$\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f}$

Where:

• $s_o$ is the object distance (distance from object to lens)
• $s_i$ is the image distance (distance from lens to image)
• $f$ is the focal length of the lens

This equation allows you to calculate any one of these values if you know the other two. For example, if you know an object is placed 20 cm from a lens with a focal length of 10 cm, you can find where the image will form:

$\frac{1}{20}+\frac{1}{s_i}=\frac{1}{10}$ $\frac{1}{s_i}=\frac{1}{10}-\frac{1}{20}=\frac{2-1}{20}=\frac{1}{20}$ $s_i=20 \text{ cm}$

Lens Focal Point Conventions

Understanding sign conventions is crucial for correctly applying the thin-lens equation and interpreting results.

• For convex lenses, the focal length is positive; for concave lenses, it's negative
• Object distances ($s_o$) are positive when the object is on the side of the lens where light originates (the "real side")
• Image distances ($s_i$) are positive when the image forms on the opposite side from where light enters
• Negative image distances indicate virtual images (on the same side as the incoming light)
• When calculating with the thin-lens equation, always maintain these sign conventions

For example, a concave lens with a focal length of -15 cm will always produce a virtual image, indicated by a negative value for $s_i$ when you solve the thin-lens equation.

Image Magnification

The magnification of an image tells us how much larger or smaller the image is compared to the object, and whether it's upright or inverted.

$|M|=\left|\frac{h_i}{h_o}\right|=\left|\frac{s_i}{s_o}\right|$

Where:

• $M$ is the magnification
• $h_i$ is the image height
• $h_o$ is the object height
• $s_i$ is the image distance
• $s_o$ is the object distance

The sign of $M$ provides important information:

• Positive $M$ means the image is upright (same orientation as the object)
• Negative $M$ means the image is inverted (upside-down)
• $|M| > 1$ means the image is larger than the object
• $|M| < 1$ means the image is smaller than the object

For instance, if $M = -2$, the image is inverted and twice as large as the object. If $M = 0.5$, the image is upright and half the size of the object.

Ray Diagrams for Lenses

Ray diagrams are graphical tools that help visualize how lenses form images. By tracing specific light rays through a lens, we can determine the location, size, orientation, and type of image formed.

To construct a ray diagram, we typically trace three key rays from the top of the object:

1. A ray parallel to the principal axis - After passing through the lens, this ray:

   • For a convex lens: passes through the far-side focal point
   • For a concave lens: appears to come from the near-side focal point

2. A ray through the center of the lens - This ray passes straight through without bending

3. A ray through (or toward) a focal point - After passing through the lens, this ray:

   • For a convex lens: a ray through the near-side focal point emerges parallel to the principal axis
   • For a concave lens: a ray aimed at the far-side focal point emerges parallel to the principal axis

Where these rays intersect (or appear to intersect when extended backward), they form the image. The diagram immediately reveals:

• Whether the image is real (rays actually intersect) or virtual (rays appear to intersect when extended backward)
• Whether the image is upright or inverted
• The relative size of the image compared to the object

Ray diagrams are particularly useful for quickly predicting image characteristics without calculations, though they're less precise than using the thin-lens equation.

Practice Problem 1: Thin-Lens Equation

Solution

Let's use the thin-lens equation to find the image position:

$\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f}$

Given:

• Object distance $s_o = 15$ cm
• Focal length $f = 10$ cm (positive because it's a convex lens)

Rearranging to solve for $s_i$:

$\frac{1}{s_i}=\frac{1}{f}-\frac{1}{s_o}=\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30}$

Therefore, $s_i = 30$ cm

The image is 30 cm from the lens on the opposite side from the object. Since $s_i$ is positive, the image is real.

To determine magnification: $M = -\frac{s_i}{s_o} = -\frac{30}{15} = -2$

The negative sign indicates the image is inverted, and the magnitude of 2 means the image is twice the size of the object.

Characteristics: The image is real, inverted, and magnified by a factor of 2.

Practice Problem 2: Ray Diagrams and Image Formation

Solution

First, let's use the thin-lens equation:

$\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f}$

Given:

• Object distance $s_o = 5$ cm
• Focal length $f = 10$ cm

Rearranging to solve for $s_i$:

$\frac{1}{s_i}=\frac{1}{f}-\frac{1}{s_o}=\frac{1}{10}-\frac{1}{5}=\frac{1-2}{10}=\frac{-1}{10}$

Therefore, $s_i = -10$ cm

The negative value for $s_i$ indicates that the image is virtual and forms on the same side of the lens as the object.

For magnification: $M = -\frac{s_i}{s_o} = -\frac{-10}{5} = 2$

The positive value indicates the image is upright, and the magnitude of 2 means the image is twice the size of the object.

Ray diagram analysis would show:

1. A ray parallel to the principal axis would refract through the far-side focal point
2. A ray through the center of the lens would continue straight
3. A ray aimed toward the far-side focal point would refract parallel to the principal axis

These rays would not actually intersect but would appear to diverge from a point 10 cm behind the object (on the same side as the object). This confirms our calculation that the image is virtual, upright, and magnified.

Characteristics: The image is virtual, upright, and magnified by a factor of 2.