IB MYP 4-5 Physics- Lenses- Study Notes - New Syllabus
Lenses
Lenses
A lens is a transparent piece of glass or plastic that refracts light rays to form an image.
Types of Lenses:
- Convex Lens (Converging Lens): Thicker at the center than at the edges. It brings parallel rays of light together (converges them) to a point called the principal focus.
- Concave Lens (Diverging Lens): Thinner at the center than at the edges. It spreads parallel rays of light outward (diverges them) as if they are coming from a point called the principal focus.
Key Terms:
- Principal Axis: A straight line passing through the optical center and the centers of curvature of the lens surfaces.
- Optical Center (O): A point on the lens through which light passes without deviation.
- Principal Focus (F): The point where parallel rays of light either converge (convex lens) or appear to diverge from (concave lens).
- Focal Length (f): The distance between the optical center and the principal focus.
Lens Formula:
\(\dfrac{1}{f} = \dfrac{1}{v} – \dfrac{1}{u}\)
where \(f\) = focal length, \(v\) = image distance, \(u\) = object distance (sign convention applied).
Magnification (M):
\( M =- \dfrac{\text{Image height}}{\text{Object height}} =- \dfrac{v}{u} \)
Ray Diagrams for Lenses:
For converging lenses:
- A ray parallel to the principal axis passes through the focal point after refraction.
-
- A ray passing through the center of the lens continues in a straight line.
- A ray through the focal point emerges parallel to the axis.
For concave lenses:
- Parallel ray → appears to diverge from the focus.
- Ray through optical center → passes undeviated.
Applications of Lenses:
- Convex lenses: magnifying glasses, cameras, microscopes, human eye.
- Concave lenses: spectacles for short-sightedness, peepholes in doors.
Example:
An object is placed 30 cm from a convex lens of focal length 15 cm. Find the image distance and state the nature of the image.
▶️ Answer/Explanation
Lens formula: \(\dfrac{1}{f} = \dfrac{1}{v} – \dfrac{1}{u}\) \(\dfrac{1}{15} = \dfrac{1}{v} – \dfrac{1}{(-30)}\)
\(\dfrac{1}{15} = \dfrac{1}{v} + \dfrac{1}{30}\) \(\dfrac{1}{v} = \dfrac{1}{15} – \dfrac{1}{30} = \dfrac{1}{30}\) \( v = 30 \, \text{cm} \).
Example:
An object is placed 20 cm from a concave lens of focal length 15 cm. Find the image distance.
▶️ Answer/Explanation
Lens formula: \(\dfrac{1}{f} = \dfrac{1}{v} – \dfrac{1}{u}\) Here, \(f = -15 \, \text{cm}, \, u = -20 \, \text{cm}\). \(\dfrac{1}{-15} = \dfrac{1}{v} – \dfrac{1}{-20}\) \(\dfrac{1}{-15} = \dfrac{1}{v} + \dfrac{1}{20}\) \(\dfrac{1}{v} = -\dfrac{1}{15} – \dfrac{1}{20} = -\dfrac{7}{60}\) \( v = -8.57 \, \text{cm} \).
The image is virtual, upright, and smaller.
Final Answer: \(\boxed{v = -8.6 \, \text{cm}, \, \text{Virtual, Upright, Diminished}}\)
Example:
A magnifying glass is a convex lens of focal length 10 cm. An object is placed 6 cm from the lens. Calculate the magnification.
▶️ Answer/Explanation
Lens formula: \(\dfrac{1}{f} = \dfrac{1}{v} – \dfrac{1}{u}\) \(\dfrac{1}{10} = \dfrac{1}{v} – \dfrac{1}{(-6)}\) \(\dfrac{1}{10} = \dfrac{1}{v} + \dfrac{1}{6}\) \(\dfrac{1}{v} = \dfrac{1}{10} – \dfrac{1}{6} = -\dfrac{2}{15}\) \( v = -7.5 \, \text{cm} \).
Magnification: \( M = \dfrac{v}{u} = \dfrac{-7.5}{-6} = 1.25 \).
Final Answer: \(\boxed{M = 1.25, \, \text{Virtual, Upright, Enlarged}}