What is magnification by a lens?

Linear magnification (m) by a lens is the ratio of the height of the image (hi) to the height of the object (ho). It describes how much larger or smaller the image is compared to the object.

What is the formula for lens magnification?

The magnification formula is m = hi / ho = v / u, where hi is the image height, ho is the object height, v is the image distance, and u is the object distance (using sign conventions).

How does the lens magnification formula differ from the mirror magnification formula?

For a lens, magnification is m = v / u (positive ratio). For a mirror, magnification is m = -v / u (negative ratio).

What does a positive magnification value mean?

A positive magnification (m > 0) indicates that the image is virtual and upright (right-side up) relative to the object. This occurs for convex lenses when u < f, and always for concave lenses.

What is the condition for an enlarged image (magnification > 1)?

An image is enlarged if the absolute value of magnification is greater than 1 (|m| > 1). For a convex lens, this occurs when the object is placed between F and 2F (real image) or inside F (virtual image).

What is the condition for a diminished image (magnification < 1)?

An image is diminished if the absolute value of magnification is less than 1 (|m| < 1). This occurs when the object is beyond 2F for a convex lens, and always at any distance for a concave lens.

What does a magnification of exactly -1 represent?

A magnification of -1 means the image is real, inverted, and the exact same size as the object (|m| = 1). This happens when a real object is placed exactly at twice the focal length of a convex lens (u = -2f, v = +2f).

What does a magnification of exactly +1 represent?

A magnification of +1 would mean a virtual, upright image of the exact same size. However, for a single lens, this is a boundary condition that only occurs when the object is in contact with the lens (u = 0, v = 0).

How do camera zoom lenses change magnification?

A camera zoom lens shifts internal glass elements to change the effective focal length (f) of the lens system. According to the lens formula, changing f alters the image distance v for a given u, thereby resizing the image on the sensor (m = v/u).

Why does a magnifying glass only work when u < f?

When the object is inside the focal length (u u in magnitude, so m = v/u is positive and greater than 1, creating an upright, enlarged view. If u > f, it forms an inverted image, which cannot be read easily.

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Ray Optics & Sizing

Magnification by Lens

Q: What is the condition for a diminished image (magnification < 1)?

An image is diminished if the absolute value of magnification is less than 1 (|m| < 1). This occurs when the object is beyond 2F for a convex lens, and always at any distance for a concave lens.

Q: Can a concave lens ever produce a magnification greater than 1?

No. A concave (diverging) lens only forms virtual, upright, and diminished images. The image distance v is always negative and smaller in magnitude than the object distance u, so m = v/u is always between 0 and +1.

Q: What does a magnification of exactly -1 represent?

A magnification of -1 means the image is real, inverted, and the exact same size as the object (|m| = 1). This happens when a real object is placed exactly at twice the focal length of a convex lens (u = -2f, v = +2f).

Q: What does a magnification of exactly +1 represent?

A magnification of +1 would mean a virtual, upright image of the exact same size. However, for a single lens, this is a boundary condition that only occurs when the object is in contact with the lens (u = 0, v = 0).

Q: How do camera zoom lenses change magnification?

A camera zoom lens shifts internal glass elements to change the effective focal length (f) of the lens system. According to the lens formula, changing f alters the image distance v for a given u, thereby resizing the image on the sensor (m = v/u).

Explore how thin lenses scale and orient optical images. Toggle between a magnifying glass workspace, a camera zoom simulator, and an optics bench height laboratory to see how image height and object height relate dynamically.

Linear Magnification Laboratory ⓘ

Drag the heights and positions to scale the virtual/real image. Inspect how sign and size change together.

Ready

Live Telemetry

m = h_i / h_o = v / u ⇒ m = -6.0 / 3.0 = -2.00

Magnification (m): +2.00 ×
Object Height (h_o): 3.0 cm
Image Height (h_i): +6.0 cm
Distances ratio (v/u): -30.0 / -15.0
Image Scaling: Enlarged
Orientation: Upright & Virtual

What is Lens Magnification?

Refraction does not merely shift where light meets; it also alters the perceived size and orientation of objects. The factor by which a lens expands or shrinks a subject is described as linear magnification (or lateral magnification).

It is defined fundamentally as the ratio of the final image height to the initial object height. Due to similar triangles formed by principal optical rays intersecting at the optical center, this ratio also equals the ratio of image distance to object distance.

Interpretation of Magnification (m)

The sign and magnitude of the magnification value m give a complete physical description of the image:

1. Direction & Orientation (Sign of m)

m is Negative (−): The image is inverted (upside down) and real. This forms on the opposite side of a convex lens when u > f.
m is Positive (+): The image is upright (right-side up) and virtual. This forms on the same side as the object (convex u < f, and all concave setups).

2. Sizing Scale (Magnitude |m|)

|m| > 1: The image is enlarged (larger than the object).
|m| < 1: The image is diminished (smaller than the object).
|m| = 1: The image is the same size as the object (occurs at u = -2f for convex).

Sizing Mathematical Relations

Linear magnification ratios are governed by the following formulas:

Magnification Ratio

m = h_i / h_o = v / u

Image Sizing

h_i = m × h_o

Where:
• h_o is object height (conventionally positive)
• h_i is image height (+ for upright, − for inverted)
• u is object distance (negative coordinate on left)
• v is image distance (+ on right, − on left)

Lens vs. Mirror Magnification

A side-by-side comparison of linear sizing rules between transmissive lenses and reflective mirrors:

Feature	Spherical Lenses	Spherical Mirrors
Primary Formula	m = v / u	m = −v / u
Upright Image (m > 0)	Virtual (same side, v < 0)	Virtual (behind mirror, v > 0)
Inverted Image (m < 0)	Real (opposite side, v > 0)	Real (in front of mirror, v < 0)
Focal Equation	1/f = 1/v − 1/u	1/f = 1/v + 1/u
Diverging System	Concave (diminished virtual)	Convex (diminished virtual)

Real-World Magnification

Lenses regulate image sizes in devices we use daily:

Camera Zoom Lenses: By sliding convex and concave glass elements inside the barrel, the camera shifts the focal length (f). This changes the image distance (v), magnifying the image details onto the sensor.
Eyepiece in Microscopes: Microscopes combine a small-focal-length objective lens (which produces an enlarged real image) with an eyepiece lens (acting as a magnifier to blow up that real image even further).
Projector Lenses: By placing a small transparent film slide just outside the focal point of a strong convex lens (u ≈ -f), a massive real image is projected onto a screen meters away (v >> u, yielding m >> 1).

Step-by-Step Solved Problems

Practice calculations using height ratios and distance ratios. Learn to navigate the algebraic signs.

Example 1 Problem Statement

A convex lens of focal length 15 cm forms a real image of a candle flame. If the candle is placed 22.5 cm from the lens and has a height of 4.0 cm, calculate the image distance, the magnification, and the height of the image.

View Step-by-Step Sizing Solution

Identify the given parameters with sign conventions: Object height h_o = +4.0 cm, Object distance u = -22.5 cm, Focal length f = +15 cm (convex lens).
Use the lens formula to find the image distance v: 1/f = 1/v - 1/u ⇒ 1/15 = 1/v - 1/(-22.5) ⇒ 1/15 = 1/v + 1/22.5.
Solve for 1/v: 1/v = 1/15 - 1/22.5 = 1.5/22.5 - 1/22.5 = 0.5 / 22.5 = 1 / 45 ⇒ v = +45 cm. The real image forms 45 cm to the right of the lens.
Calculate linear magnification: m = v / u = 45 / (-22.5) = -2.00.
Use the magnification formula to find the image height h_i: m = h_i / h_o ⇒ -2.00 = h_i / 4.0 ⇒ h_i = -2.00 × 4.0 = -8.0 cm.
Interpret the final result: The image is formed 45 cm on the opposite side of the lens, is twice as large as the candle flame (8.0 cm high), and is inverted (indicated by the negative sign).

Final Derived Answer: Image Distance v = +45 cm, Magnification m = -2.00, Image Height h_i = -8.0 cm (inverted).

Example 2 Problem Statement

A stamp is viewed through a magnifying glass with a focal length of 10 cm. If the stamp is placed 7.5 cm from the lens and has a height of 2.0 cm, find the position, magnification, and size of the virtual image.

View Step-by-Step Sizing Solution

Identify the given values: Object height h_o = +2.0 cm, Object distance u = -7.5 cm, Focal length f = +10 cm (magnifier convex lens).
Recall the lens formula: 1/f = 1/v - 1/u.
Substitute values: 1/10 = 1/v - 1/(-7.5) ⇒ 1/10 = 1/v + 1/7.5.
Solve for 1/v: 1/v = 1/10 - 1/7.5 = 1/10 - 4/30 = 3/30 - 4/30 = -1/30 ⇒ v = -30 cm. The negative sign shows it is a virtual image on the same side as the stamp.
Calculate magnification: m = v / u = -30 / (-7.5) = +4.00.
Calculate image height: h_i = m × h_o = +4.00 × 2.0 = +8.0 cm.
Conclude the nature: The virtual image is formed 30 cm in front of the magnifier, is upright (positive height), and is magnified 4 times to a height of 8.0 cm.

Final Derived Answer: Image Position v = -30 cm, Magnification m = +4.00, Image Height h_i = +8.0 cm (upright).

Example 3 Problem Statement

An insect is placed 20 cm in front of a concave lens of focal length 20 cm. If the insect is 1.5 cm tall, calculate the magnification and the height of the image formed.

View Step-by-Step Sizing Solution

Identify the parameters: Object height h_o = +1.5 cm, Object distance u = -20 cm, Focal length f = -20 cm (concave lens).
Recall the lens formula: 1/f = 1/v - 1/u.
Substitute values: 1/(-20) = 1/v - 1/(-20) ⇒ -1/20 = 1/v + 1/20.
Solve for 1/v: 1/v = -1/20 - 1/20 = -2/20 = -1/10 ⇒ v = -10 cm.
Calculate magnification: m = v / u = -10 / (-20) = +0.50.
Calculate image height: h_i = m × h_o = +0.50 × 1.5 = +0.75 cm.
Describe nature: The image forms 10 cm in front of the lens on the same side, is virtual and upright, and is diminished to half the original size (0.75 cm tall).

Final Derived Answer: Magnification m = +0.50, Image Height h_i = +0.75 cm (upright, diminished).

Self-Check Questions

Question 1

A lens has a magnification of -0.5. Explain what this value tells you about the image's height, orientation, and nature.

Show Answer & Explanation

A magnification value of -0.5 tells us three things: (1) The negative sign indicates the image is inverted and real. (2) The magnitude is 0.5 (which is less than 1), meaning the image is diminished to half the height of the object. (3) The image forms on the opposite side of the lens relative to the object.

Question 2

Under what condition does a converging lens produce a magnification of exactly -1?

Show Answer & Explanation

A convex (converging) lens produces a magnification of exactly -1 when the object is placed at a distance of twice the focal length from the lens (u = -2f). In this position, the image distance is also v = +2f, so m = v/u = 2f/(-2f) = -1. The image is real, inverted, and of the same size.

Question 3

An object is placed 12 cm from a convex lens of focal length 8 cm. Calculate the magnification.

Show Answer & Explanation

Given: u = -12 cm, f = 8 cm. First find v: 1/v = 1/f + 1/u = 1/8 - 1/12 = (3 - 2)/24 = 1/24 ⇒ v = +24 cm. Now calculate magnification: m = v/u = 24 / (-12) = -2.00. The image is inverted, real, and twice the size of the object.

Question 4

Can a single diverging (concave) lens ever produce a magnification greater than +1? Explain why.

Show Answer & Explanation

No. A concave lens diverges light rays. For any real object, the virtual image distance v is always negative and smaller in magnitude than the object distance u (|v| < |u|). Since m = v/u and both v and u are negative, m is always positive but strictly less than 1. Thus, a concave lens only forms diminished images.

Question 5

A slide projector uses a lens to project a slide of height 3 cm onto a screen. If the screen is 3.0 meters away and the image is 90 cm tall, find the magnification and the object distance.

Show Answer & Explanation

Given: ho = 3 cm, hi = -90 cm (inverted on screen), v = +300 cm. Magnification m = hi / ho = -90 / 3 = -30. Using m = v/u: -30 = 300 / u ⇒ u = 300 / (-30) = -10 cm. The slide must be placed 10 cm in front of the projection lens.

Question 6

How does changing the height of the object (h<sub>o</sub>) affect the magnification (m) of a lens if the object's position is held constant?

Show Answer & Explanation

Magnification is independent of the object's height. It is determined solely by the ratio of image distance to object distance (m = v/u), which depends only on the object distance u and focal length f. Changing ho will proportionally change the image height hi so that the ratio hi/ho remains constant.

Linear Magnification Laboratory ⓘ

Object Name

Live Telemetry

What is Lens Magnification?

Interpretation of Magnification (m)

1. Direction & Orientation (Sign of m)

2. Sizing Scale (Magnitude |m|)

Sizing Mathematical Relations

Lens vs. Mirror Magnification

Real-World Magnification

Step-by-Step Solved Problems

Example 1 Problem Statement

Example 2 Problem Statement

Example 3 Problem Statement

Self-Check Questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

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