Lens Formula, Magnification, and Power of a Lens - Tenth Grade Science
The quantitative analysis of image formation by thin lenses relies on algebraic relations that link object distance, image distance, focal length, and optical strength. Establishing a rigorous coordinate framework through the Cartesian sign convention enables accurate calculations of image positions, orientations, and the corrective power required for ophthalmic lenses.
The New Cartesian Sign Convention for Lenses
To ensure mathematical consistency across all lens problems, distances are measured within a standard Cartesian coordinate frame where the lens optical center serves as the origin.
NEW CARTESIAN SIGN CONVENTION FOR LENSES
+y (Height Upward: Positive)
|
Incident Light Direction |
======================> |
|
Distances AGAINST Light | Distances WITH Light
(Negative -x) | (Positive +x)
---------------------------------O--------------------------------- Principal Axis
|
|
|
-y (Height Downward: Negative)
|
Reference Rules
Every measurement is taken along or perpendicular to the principal axis starting from the optical center $O$, which functions as the origin $(0, 0)$. The incident light is always assumed to travel from left to right across the diagram. All horizontal distances measured in the direction of the incident light (to the right of $O$) are defined as positive ($+x$). All horizontal distances measured opposite to the direction of incident light (to the left of $O$) are defined as negative ($-x$). Vertical heights measured upward perpendicular to the principal axis are defined as positive ($+y$). Vertical heights measured downward perpendicular to the principal axis are defined as negative ($-y$).
Application to Lens Types
Because real objects are positioned to the left of the lens to face incoming light, the object distance ($u$) is universally negative ($u < 0$) in standard single-lens configurations. A convex lens focuses parallel incident light onto its secondary focal point $F_2$ situated to the right of the lens. Therefore, the focal length of a convex lens is always positive ($f > 0$). A concave lens diverges parallel incident light such that the rays appear to originate from its virtual focus $F_1$ located to the left of the lens. Hence, the focal length of a concave lens is always negative ($f < 0$).
The Thin Lens Formula
The thin lens formula mathematically relates the object distance ($u$), the image distance ($v$), and the focal length ($f$) of a spherical lens.
Derivation Sketch and Algebraic Expression
By considering similar right triangles formed by the object, image, optical center, and principal focus in standard ray tracing, geometric proportionality yields the classical thin lens equation:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
In this formula: $u$ is the object distance from the optical center. $v$ is the image distance from the optical center. $f$ is the focal length of the lens.
GEOMETRIC DERIVATION PROPORTIONS
A
| \
h | \
| \
-------B----O------------------------- Principal Axis
| \ |
| \ h' |
| \ |
\ v
-------------------A'
Triangle ABO is similar to Triangle A'B'O: A'B' / AB = OB' / OB => -h' / h = v / -u
Using focal triangle proportionality yields: 1/v - 1/u = 1/f
Distinction from the Mirror Formula
Care must be exercised not to confuse the thin lens formula with the spherical mirror formula. The mirror formula contains a positive sign ($\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$), reflecting the fact that light reverses its direction upon reflection. In contrast, the lens formula features a negative sign ($\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$), arising because light transmits through the lens to form real images on the opposite side.
Linear Magnification
Linear magnification ($m$) quantifies the ratio of the vertical size of the formed image to the vertical size of the original object.
Height Ratio Formulation
Let $h$ represent the transverse height of the object and $h'$ represent the transverse height of the image. The linear magnification is defined as:
$$m = \frac{\text{Height of Image}}{\text{Height of Object}} = \frac{h'}{h}$$
Distance Ratio Formulation
Using triangle similarity between rays passing directly through the optical center $O$, the ratio of image height to object height directly matches the ratio of image distance to object distance:
$$m = \frac{h'}{h} = \frac{v}{u}$$
Notice that unlike spherical mirrors where magnification is given by $m = -\frac{v}{u}$, the magnification for a thin lens has a positive quotient ($m = +\frac{v}{u}$).
Physical Interpretation of Magnification Values
If $m$ is positive ($m > 0$), the image is erect (upright) relative to the object, indicating a virtual image. If $m$ is negative ($m < 0$), the image is inverted relative to the object, confirming a real image. If $|m| > 1$, the image is magnified (larger than the object). If $|m| = 1$, the image has the identical size as the object. If $|m| < 1$, the image is diminished (smaller than the object).
MAGNIFICATION SIGN AND VALUE GUIDE
Value of m Image Nature Image Orientation Size Comparison
-----------------------------------------------------------------------------
m = -2.5 Real Inverted Magnified (2.5x)
m = -1.0 Real Inverted Same size (1.0x)
m = -0.4 Real Inverted Diminished (0.4x)
m = +1.8 Virtual Erect Magnified (1.8x)
m = +0.5 Virtual Erect Diminished (0.5x)
Power of a Lens
The power of an optical lens measures its ability to bend or deviate incoming light rays. A lens that bends light strongly focuses parallel rays over a short distance, possessing high optical power.
Mathematical Definition and SI Unit
The power ($P$) of a lens is defined as the reciprocal of its focal length ($f$) measured in meters:
$$P = \frac{1}{f \text{ (in meters)}}$$
If the focal length is provided in centimeters, the equation can be written as:
$$P = \frac{100}{f \text{ (in centimeters)}}$$
The SI unit of optical lens power is the dioptre, symbolized by the uppercase letter $\text{D}$. One dioptre represents the power of a lens having a focal length of exactly one meter ($1 \text{ D} = 1 \text{ m}^{-1}$).
LENS POWER AND FOCAL LENGTH RELATION
Short Focal Length Long Focal Length
(High Curvature) (Low Curvature)
| |
-----\ | /----- -------\ | /-------
\|/ \|/
* Focus |
/|\ * Focus
-----/ | \----- /|\
-------/ | \-------
High Power (+D) Low Power (+D)
Sign of Lens Power
For a convex (converging) lens, the focal length is positive ($f > 0$), which means its optical power is strictly positive ($P > 0$). An eyeglass prescription specifying $+2.5 \text{ D}$ indicates a converging convex lens. For a concave (diverging) lens, the focal length is negative ($f < 0$), meaning its optical power is strictly negative ($P < 0$). A prescription specifying $-1.5 \text{ D}$ indicates a diverging concave lens.
Combinations of Thin Lenses
In practical optical instruments such as compound microscopes, astronomical telescopes, and modern multi-element camera lenses, several thin lenses are placed in coaxial contact to eliminate aberrations and control magnification.
Net Power and Effective Focal Length
When two thin lenses of powers $P_1$ and $P_2$ are placed in direct physical contact along a common principal axis, their combined optical power ($P$) is the algebraic sum of their individual powers:
$$P = P_1 + P_2 + P_3 + \dots$$
Expressed in terms of effective focal length ($F$):
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} + \dots$$
Net Magnification
The total magnification produced by an optical system composed of lenses in contact (or separated along a tube) is the product of their individual linear magnifications:
$$m = m_1 \times m_2 \times m_3 \times \dots$$
This multiplicative property explains how a compound microscope combining an objective lens ($m_1 = 40$) with an eyepiece ($m_2 = 10$) achieves an overall magnification of $m = 40 \times 10 = 400$.
Chapter Exercises and Solutions
Exercise 1: An object 4.0 cm tall is placed 25.0 cm in front of a convex lens having a focal length of 10.0 cm. Calculate the image distance, the height of the image, and state whether the image is real or virtual.
Solution 1: According to the Cartesian sign convention, the given values are object height $h = +4.0 \text{ cm}$, object distance $u = -25.0 \text{ cm}$, and focal length $f = +10.0 \text{ cm}$. Applying the thin lens formula gives $1/v - 1/u = 1/f$, which rearranges to $1/v = 1/f + 1/u$. Substituting the numerical parameters gives $1/v = 1/10.0 + 1/(-25.0) = 1/10.0 - 1/25.0$. Finding a common denominator of 50 yields $1/v = 5/50 - 2/50 = 3/50$, so $v = 50/3 \approx +16.67 \text{ cm}$. Because $v$ is positive, the image is formed 16.67 cm to the right of the lens, confirming it is real and inverted. To find image height, use the magnification relation $m = h'/h = v/u$, so $h' = h \times (v/u) = 4.0 \times ((50/3) / (-25.0)) = 4.0 \times (-2/3) \approx -2.67 \text{ cm}$. The formed image is real, inverted, has a height of 2.67 cm, and is situated 16.67 cm behind the lens.
Exercise 2: A concave lens has a focal length of 20 cm. At what distance should an object be placed from the lens so that it produces an image positioned 10 cm from the lens? Also compute the magnification produced by the lens.
Solution 2: For a concave lens, both the focal length and the virtual image distance are negative under the sign convention, giving $f = -20 \text{ cm}$ and $v = -10 \text{ cm}$. Rearranging the thin lens equation $1/v - 1/u = 1/f$ to isolate object distance gives $1/u = 1/v - 1/f$. Substituting the values gives $1/u = 1/(-10) - 1/(-20) = -1/10 + 1/20 = -2/20 + 1/20 = -1/20$. Inverting this fraction yields $u = -20 \text{ cm}$. The object must be positioned 20 cm in front of the concave lens. The linear magnification is given by $m = v / u = (-10) / (-20) = +0.5$. The positive magnification confirms that the image is virtual, erect, and half the height of the original object.
Exercise 3: An optometrist prescribes a corrective combination consisting of two thin lenses placed in contact: a convex lens of focal length $+25 \text{ cm}$ and a concave lens of focal length $-50 \text{ cm}$. Calculate the individual powers of each lens, the total combined power of the pair, and the net effective focal length of the system.
Solution 3: First convert each focal length into meters: $f_1 = +25 \text{ cm} = +0.25 \text{ m}$ and $f_2 = -50 \text{ cm} = -0.50 \text{ m}$. Using the power relation $P = 1 / f$, the individual powers are $P_1 = 1 / (+0.25) = +4.0 \text{ D}$ and $P_2 = 1 / (-0.50) = -2.0 \text{ D}$. The total power of lenses in contact is their algebraic sum, $P = P_1 + P_2 = +4.0 \text{ D} + (-2.0 \text{ D}) = +2.0 \text{ D}$. The effective focal length of the combined pair is $F = 1 / P = 1 / (+2.0 \text{ D}) = +0.50 \text{ m} = +50 \text{ cm}$. The combined optical system functions as a net converging lens with a power of $+2.0 \text{ D}$ and an effective focal length of $+50 \text{ cm}$.