The Physics of Lenses and Mirrors
Geometric optics describes how light propagates in terms of rays. When light hits a surface, it can be reflected (mirrors) or refracted (lenses). The fundamental equations of optics allow us to predict where an image will form and what its characteristics will be.
The Mirror and Lens Equations
While similar, the equations for mirrors and thin lenses differ by a sign convention:
Thin Lens Equation: 1/f = 1/v - 1/u
Where:
- f is the focal length.
- u is the object distance (distance from object to lens/mirror).
- v is the image distance (distance from image to lens/mirror).
Magnification
Magnification (m) describes the ratio of the image height to the object height:
- |m| > 1: The image is enlarged.
- |m| < 1: The image is diminished.
- m is positive: The image is upright (virtual).
- m is negative: The image is inverted (real).
Sign Convention (The Cartesian System)
A consistent sign convention is vital for solving optics problems:
- u (Object Distance): Always positive for real objects.
- f (Focal Length): Positive for Converging (Convex Lens / Concave Mirror). Negative for Diverging (Concave Lens / Convex Mirror).
- v (Image Distance): Positive for Real images (formed on the opposite side of a lens, or same side of a mirror). Negative for Virtual images.
Deep Dive: Worked Examples
โ Example 1: Concave Mirror (Real Image)
An object is placed 30 cm in front of a concave mirror with a focal length of 20 cm. Find the image position and nature.
Formula: 1/v = 1/f - 1/u
Solution:
1/v = 1/20 - 1/30 = (3 - 2) / 60 = 1/60
v = 60 cm
Nature: v is positive, so it is a Real and Inverted image.
โ Example 2: Convex Mirror (Virtual Image)
A car's side mirror is a convex mirror with a focal length of -50 cm. If another car is 10 m away, what is the magnification?
Find v: 1/v = 1/(-50) - 1/1000 = -20/1000 - 1/1000 = -21/1000
v = -47.6 cm
Magnification: m = -v / u = -(-47.6) / 1000
m = +0.0476 (Virtual, Upright, and highly diminished)
โ Example 3: Convex Lens (Magnifying Glass)
A stamp collector uses a convex lens of focal length 10 cm. He holds it 8 cm from a stamp. Find the magnification.
Formula: 1/v = 1/f + 1/u (Note: for lens 1/v - 1/u = 1/f)
Solution:
1/v = 1/10 + (-1/8) ? No, 1/v = 1/f + 1/u = 1/10 + 1/(-8) is wrong sign convention.
Using 1/f = 1/v - 1/u: 1/10 = 1/v - 1/(-8) -> 1/v = 1/10 - 1/8 = -1/40
v = -40 cm (Virtual image)
Magnification: m = v/u (for lens) = -40 / -8 = +5.0
โ Example 4: Concave Lens (Diverging)
A diverging lens has a focal length of 15 cm. If an object is placed 30 cm away, where is the image?
Formula: 1/v = 1/f + 1/u (Using 1/v - 1/u = 1/f convention where u is negative)
Solution:
1/v = 1/(-15) + 1/(-30) = -2/30 - 1/30 = -3/30
v = -10 cm
Nature: Virtual, Upright and diminished (formed 10 cm in front of lens).
โ Example 5: Lens Power
A person wears corrective glasses with a power of -2.5 Dioptres. What is the focal length and type of lens?
Formula: f = 1 / P (f in meters)
Solution:
f = 1 / (-2.5) = -0.4 m
f = -40 cm
Since f is negative, it is a Concave (Diverging) lens used to correct myopia.