Mirror Formula & Linear Magnification - Tenth Grade Science
While geometric ray tracing provides qualitative visualization of image formation by spherical mirrors, optical engineering and applied physics require quantitative mathematical relationships to calculate image locations, focal parameters, and image dimensions. The foundational equation of spherical mirror optics is the Mirror Formula, which relates the object distance, the image distance, and the focal length of a spherical mirror in a single algebraic expression. Paired with the concept of linear magnification, which quantifies the proportional enlargement or reduction of an image relative to its object, these mathematical tools enable the design of optical systems. In this chapter, you will investigate the geometric derivation of the mirror formula using similar triangles, derive the linear magnification equations, apply the New Cartesian Sign Convention to numerical problems, and analyze worked computational examples across concave and convex mirrors.
Geometric Derivation of the Mirror Formula
The Mirror Formula is an algebraic equation that relates the object distance ($u$), the image distance ($v$), and the focal length ($f$) of a spherical mirror:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
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| GEOMETRIC DERIVATION OF THE MIRROR FORMULA |
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| A |
| |\ |
| | \ |
| | \ |
| | \ |
| Principal | \ |
| Axis -----B-----C---------A'--------F---------P----------------------- |
| \ /| | |
| \ / | | |
| \ / | | |
| x B' | |
| / \ | |
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| Object AB placed beyond C; real inverted image A'B' formed between C and F. |
| Two sets of equiangular (similar) right-angled triangles are compared: |
| Set 1: Triangle ABC and Triangle A'B'C are similar. |
| Set 2: Triangle ABP and Triangle A'B'P are similar. |
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Derivation Steps Using Similar Triangles
Consider a concave mirror of small aperture with pole $P$, principal focus $F$, and center of curvature $C$. An object $AB$ of height $h$ is placed perpendicular to the principal axis beyond $C$, forming a real, inverted image $A'B'$ of height $h'$ between $C$ and $F$:
Step 1: Compare Triangles $\Delta ABC$ and $\Delta A'B'C$: Both triangles share equal vertically opposite angles at $C$ ($\angle ACB = \angle A'CB'$), and both have right angles at the principal axis ($\angle ABC = \angle A'B'C = 90^\circ$). Because their angles are identical, the triangles are similar ($\Delta ABC \sim \Delta A'B'C$). The ratio of their corresponding sides is equal:
$$\frac{AB}{A'B'} = \frac{BC}{B'C} \quad \text{--- (Equation 1)}$$
Step 2: Compare Triangles $\Delta ABP$ and $\Delta A'B'P$: Ray $AP$ strikes the pole at angle $i$ and reflects as ray $PA'$ at angle $r$, where $\angle APB = \angle A'PB'$. Both triangles have right angles at the principal axis ($\angle ABP = \angle A'B'P = 90^\circ$). Therefore, $\Delta ABP \sim \Delta A'B'P$, and the ratio of their corresponding sides is:
$$\frac{AB}{A'B'} = \frac{BP}{B'P} \quad \text{--- (Equation 2)}$$
Step 3: Equating the Geometric Ratios: Equating Equation 1 and Equation 2 gives:
$$\frac{BC}{B'C} = \frac{BP}{B'P} \quad \text{--- (Equation 3)}$$
Step 4: Expressing Distances from the Pole ($P$): Express all line segments along the principal axis in terms of distances measured from the pole $P$: $BP$ is the object distance: $BP = -u$ $B'P$ is the image distance: $B'P = -v$ $CP$ is the radius of curvature: $CP = -R = -2f$ Segment $BC = BP - CP = (-u) - (-R) = -u + R = -u + 2f$ Segment $B'C = CP - B'P = (-R) - (-v) = -R + v = -2f + v$
Substituting these into Equation 3:
$$\frac{-u + 2f}{-2f + v} = \frac{-u}{-v} = \frac{u}{v}$$
Cross-multiplying gives:
$$v(-u + 2f) = u(-2f + v) \implies -uv + 2vf = -2uf + uv$$
Rearranging terms:
$$2vf + 2uf = 2uv$$
Divide every term by 2:
$$vf + uf = uv$$
Step 5: Dividing by $uvf$: Divide both sides by the product $u \cdot v \cdot f$:
$$\frac{vf}{uvf} + \frac{uf}{uvf} = \frac{uv}{uvf} \implies \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$$
This yields the standard Mirror Formula:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
This equation is valid for both concave and convex spherical mirrors, and for real and virtual images, provided the New Cartesian Sign Convention is applied to all quantities ($u, v, f$).
Linear Magnification: Principles and Formulas
Linear magnification ($m$) produced by a spherical mirror is defined as the ratio of the height of the image ($h'$) to the height of the object ($h$).
$$m = \frac{\text{Height of Image }(h')}{\text{Height of Object }(h)}$$
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| LINEAR MAGNIFICATION RELATIONSHIPS |
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| h' - v |
| m = ---- = ------- |
| h u |
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| MAGNIFICATION SIGN INTERPRETATION: |
| - If m is NEGATIVE (m < 0): Image is REAL and INVERTED (h' is negative). |
| - If m is POSITIVE (m > 0): Image is VIRTUAL and ERECT (h' is positive). |
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| MAGNIFICATION MAGNITUDE INTERPRETATION (|m|): |
| - If |m| > 1: Image is MAGNIFIED (Enlarged, h' > h). |
| - If |m| = 1: Image is SAME SIZE as object (h' = h, e.g., object at C). |
| - If |m| < 1: Image is DIMINISHED (Reduced, h' < h). |
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Derivation of the Magnification Formula in Terms of $u$ and $v$
From the similar triangles $\Delta ABP \sim \Delta A'B'P$ in the derivation:
$$\frac{A'B'}{AB} = \frac{B'P}{BP}$$
Applying the New Cartesian Sign Convention: Object height is erect above the axis: $AB = +h$ Image height is inverted below the axis: $A'B' = -h'$ Object distance: $BP = -u$ Image distance: $B'P = -v$
Substituting these signed values gives:
$$\frac{-h'}{h} = \frac{-v}{-u} = \frac{v}{u} \implies \frac{h'}{h} = -\frac{v}{u}$$
Since $m = h'/h$, we obtain:
$$m = -\frac{v}{u}$$
Alternative Formulations of Magnification
Using the mirror formula $1/v + 1/u = 1/f$, magnification can be expressed in terms of focal length:
In terms of $f$ and $u$: Multiply the mirror formula by $u$:
$$\frac{u}{v} + 1 = \frac{u}{f} \implies \frac{u}{v} = \frac{u}{f} - 1 = \frac{u - f}{f} \implies -\frac{v}{u} = \frac{f}{f - u}$$
$$m = \frac{f}{f - u}$$
In terms of $f$ and $v$: Multiply the mirror formula by $v$:
$$1 + \frac{v}{u} = \frac{v}{f} \implies \frac{v}{u} = \frac{v}{f} - 1 = \frac{v - f}{f} \implies -\frac{v}{u} = -\left(\frac{v - f}{f}\right) = \frac{f - v}{f}$$
$$m = \frac{f - v}{f}$$
Quantitative Numerical Problem-Solving Protocol
Solving numerical problems in spherical mirror optics requires following a systematic protocol to avoid sign errors.
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| NUMERICAL SOLVING ALGORITHMIC PROTOCOL |
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| Step 1: Identify mirror geometry (Concave -> f < 0; Convex -> f > 0). |
| Step 2: Assign Cartesian signs to all given parameters (u is ALWAYS negative). |
| Step 3: State the Mirror Formula and isolate the unknown algebraic variable. |
| Step 4: Substitute signed values with common dimensional units (cm or m). |
| Step 5: Calculate magnification (m = -v/u) and interpret the sign and size: |
| Sign of v (- front/real, + behind/virtual); Sign of m (- inv, + erect). |
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Worked Computational Example 1: Concave Mirror
Problem: An object $4.0\text{ cm}$ in height is placed at a distance of $25.0\text{ cm}$ in front of a concave mirror of focal length $15.0\text{ cm}$. Find the position, nature, and height of the image formed.
Solution: Given data with Cartesian signs: Concave mirror: focal length $f = -15.0\text{ cm}$ Object distance: $u = -25.0\text{ cm}$ Object height: $h = +4.0\text{ cm}$ Unknowns: image distance ($v$) and image height ($h'$)
Applying the Mirror Formula:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$$
Substitute the signed values:
$$\frac{1}{v} = \frac{1}{-15.0} - \frac{1}{-25.0} = -\frac{1}{15.0} + \frac{1}{25.0}$$
The least common multiple of 15 and 25 is 75:
$$\frac{1}{v} = \frac{-5 + 3}{75} = \frac{-2}{75} \implies v = -\frac{75}{2} = -37.5\text{ cm}$$
Calculate linear magnification ($m$):
$$m = -\frac{v}{u} = -\frac{-37.5}{-25.0} = -(+1.5) = -1.5$$
Calculate image height ($h'$):
$$m = \frac{h'}{h} \implies h' = m \cdot h = (-1.5) \times 4.0\text{ cm} = -6.0\text{ cm}$$
Interpretation: Image distance $v = -37.5\text{ cm}$: The negative sign indicates the image is formed $37.5\text{ cm}$ in front of the mirror (on the same side as the object); therefore, it is a real image. Magnification $m = -1.5$ and $h' = -6.0\text{ cm}$: The negative sign indicates the image is inverted. The absolute magnitude $|m| = 1.5 > 1$ indicates the image is magnified (enlarged to $6.0\text{ cm}$ compared to the $4.0\text{ cm}$ object).
Worked Computational Example 2: Convex Mirror
Problem: A convex mirror used for rearview surveillance on an automobile has a radius of curvature of $2.0\text{ m}$. If a bus is located at $5.0\text{ m}$ from the mirror, find the position, nature, and magnification of the image of the bus.
Solution: Given data with Cartesian signs: Convex mirror: radius of curvature $R = +2.0\text{ m}$ Focal length: $f = R/2 = +2.0 / 2 = +1.0\text{ m}$ Object distance (bus): $u = -5.0\text{ m}$ Unknowns: image distance ($v$) and magnification ($m$)
Applying the Mirror Formula:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$$
Substitute the signed values:
$$\frac{1}{v} = \frac{1}{+1.0} - \frac{1}{-5.0} = \frac{1}{1.0} + \frac{1}{5.0} = \frac{5 + 1}{5} = \frac{6}{5}$$
Inverting yields:
$$v = +\frac{5}{6}\text{ m} \approx +0.833\text{ m} \quad (+83.3\text{ cm})$$
Calculate linear magnification ($m$):
$$m = -\frac{v}{u} = -\frac{+5/6}{-5} = -\left(-\frac{5}{6 \times 5}\right) = +\frac{1}{6} \approx +0.167$$
Interpretation: Image distance $v = +0.833\text{ m}$: The positive sign indicates the image is formed behind the mirror at a distance of $0.833\text{ meters}$ ($83.3\text{ cm}$) from the pole. Magnification $m = +0.167$: The positive sign indicates the image is virtual and erect. The magnitude $|m| = 0.167 < 1$ indicates the image is diminished to approximately one-sixth the size of the bus.
Chapter Exercises
Exercise 1: An object $2.0\text{ cm}$ high is placed at a distance of $12.0\text{ cm}$ in front of a concave mirror. It forms a real image that is $4.0\text{ cm}$ high. Calculate the magnification, determine the image distance, and calculate the focal length of the concave mirror.
Exercise 2: A convex mirror with a focal length of $20\text{ cm}$ forms an image of an object that is one-fourth the size of the object. Calculate the distance at which the object is placed in front of the mirror, and determine the exact position of the image behind the mirror.
Exercise 3: Derive the formula for linear magnification $m = f / (f - u)$ directly from the standard mirror formula $1/v + 1/u = 1/f$, showing every intermediate algebraic step, and explain why a plane mirror has a magnification of exactly $m = +1$.
Solutions and Explanations
Solution 1: The object height is $h = +2.0\text{ cm}$. It forms a real image that is $4.0\text{ cm}$ high; because real images are always inverted, the image height carries a negative sign: $h' = -4.0\text{ cm}$. The object distance is $u = -12.0\text{ cm}$. First, calculate linear magnification: $m = h' / h = -4.0\text{ cm} / 2.0\text{ cm} = -2.0$. Second, use the magnification formula $m = -v/u$ to find the image distance: $-2.0 = -v / (-12.0) \implies -2.0 = v / 12.0 \implies v = -2.0 \times 12.0 = -24.0\text{ cm}$. The image is formed $24.0\text{ cm}$ in front of the mirror. Third, apply the mirror formula to find the focal length ($f$): $1/f = 1/v + 1/u = 1/(-24.0) + 1/(-12.0) = -1/24.0 - 2/24.0 = -3/24.0 = -1/8.0\text{ cm}^{-1}$. Inverting gives $f = -8.0\text{ cm}$. The focal length of the concave mirror is $8.0\text{ cm}$.
Solution 2: For the convex mirror, the focal length is positive: $f = +20\text{ cm}$. The image formed by a convex mirror is always virtual and erect, meaning the magnification is positive: $m = +1/4 = +0.25$. Using the magnification formula: $m = -v/u \implies +1/4 = -v/u \implies u = -4v$, or $v = -u/4$. Substitute this into the mirror formula $1/v + 1/u = 1/f$: $1/(-u/4) + 1/u = 1/20 \implies -4/u + 1/u = 1/20 \implies -3/u = 1/20$. Cross-multiplying gives $u = -3 \times 20 = -60\text{ cm}$. The object is placed $60\text{ cm}$ in front of the convex mirror. Now calculate the image position: $v = -u/4 = -(-60\text{ cm}) / 4 = +15\text{ cm}$. The image is formed $15\text{ cm}$ behind the convex mirror.
Solution 3: Start with the standard mirror formula: $1/v + 1/u = 1/f$. To express magnification ($m = -v/u$) in terms of $f$ and $u$, multiply every term of the equation by the object distance $u$: $u(1/v) + u(1/u) = u(1/f) \implies u/v + 1 = u/f$. Isolate the term $u/v$: $u/v = u/f - 1 = (u - f) / f$. Take the reciprocal of both sides: $v/u = f / (u - f)$. Multiply both sides by $-1$ to obtain the definition of linear magnification ($m = -v/u$): $m = -v/u = -[f / (u - f)] = f / [-(u - f)] = f / (f - u)$. This derives $m = f / (f - u)$. A plane mirror can be considered a spherical mirror with an infinite radius of curvature ($R = \infty$) and infinite focal length ($f = \infty$). For a plane mirror, the image distance equals the negative of the object distance ($v = -u$), and the image is virtual and erect with an image height identical to the object height ($h' = h$). Substituting these values into the magnification definition gives $m = h'/h = +h/h = +1$ (or $m = -(-u)/u = +u/u = +1$). The positive sign indicates the image is erect (upright), and the numerical value of 1 indicates the image is the exact same size as the object.