13.4 The Lens

13.4 The Lens

Overview

  • Section 13.4 covers the fundamental principles of lenses, specifically focusing on the characteristics and formation of images using two methods: ray diagrams and algebra.

Key Methods for Determining Image Characteristics

  • There are two primary ways to understand the characteristics of images formed by lenses:

    • Ray Diagrams: A graphical representation to show the pathway of light rays through a lens.

    • Algebra: Mathematical equations that relate various properties of the lens and the images formed.

Thin Lens Equation

Definition

  • The Thin Lens Equation is a crucial formula used to analyze the relationship among the focal length (f), the object distance (do), and the image distance (di).

(1)rac1f=rac1d<em>o+rac1d</em>i(1) rac{1}{f} = rac{1}{d<em>o} + rac{1}{d</em>i}

Sign Conventions

  • The following conventions apply to the quantities involved in the Lens Equation:

    • Object Distance (d_o): Always positive.

    • Focal Length (f):

    • Positive if the lens is converging (convex lens).

    • Negative if the lens is diverging (concave lens).

    • Image Distance (d_i):

    • Positive if the image is real (formed on the opposite side from the object).

    • Negative if the image is virtual (same side as the object).

Magnification Equation

Definition

  • The magnification equation is used to calculate the magnification (M) of the image produced by a lens.

(2)M=rach<em>ih</em>o(2) M = rac{h<em>i}{h</em>o}

Where:

  • h_i = height of the image

  • h_o = height of the object

Sign Conventions for Magnification

  • The sign conventions for the magnification equation are as follows:

    • Height of Object (h_o):

    • Positive when the object is upright.

    • Negative when the object is inverted.

    • Height of Image (h_i):

    • Positive when the image is upright.

    • Negative when the image is inverted.

    • Magnification (M):

    • Positive when the image is upright.

    • Negative when the image is inverted.

Summary of Sign Conventions for Lenses

Variable

Description

Positive

Negative

d_o

Object Distance

Always

Never

d_i

Image Distance

Real image (on opposite side of lens from object)

Virtual image (on same side of lens as object)

h_o

Height of Object

Measured upward

Measured downward

h_i

Height of Image

Measured upward

Measured downward

f

Focal Length

Converging lens

Diverging lens

M

Magnification

Upright image

Inverted image

Examples

Example 1: Using the Thin Lens Equation for a Converging Lens

  • Scenario: A converging lens with a focal length of 17 cm has a candle located at 48 cm from the lens.

  • Analysis: Using the thin lens equation:

    • Type of image formed: Real

    • Location: Approximately 26 cm from the lens, on the opposite side of the object.

Example 2: Using the Thin Lens Equation for a Diverging Lens

  • Scenario: A diverging lens with a focal length of 29 cm has a virtual image of a marble located 13 cm in front of the lens.

  • Analysis: The marble is located 23 cm from the lens on the same side.

Example 3: Finding the Magnification of a Converging Lens

  • Scenario: A toy with a height of 8.4 cm creates an inverted, real image of height 23 cm.

  • Calculation:

    • Magnification (

    • Result: Magnification calculated as:
      M=rach<em>ih</em>o=rac238.4ext(approximately)M = rac{h<em>i}{h</em>o} = rac{23}{8.4} ext{ (approximately) }

Example 4: Locating the Image

  • Scenario: A small toy block placed 7.2 cm in front of a lens produces an upright, virtual image with a magnification of 3.2.

  • Analysis:

    • The image location can be determined using including the magnification equation and the given object distance.

Additional Examples

Example 5: Type of Image Formation

  • Scenario: A 4.00-cm tall light bulb is placed at 8.30 cm from a double convex lens with a focal length of 15.2 cm.

  • Analysis: Determine the type of image formed and its location using the thin lens equation.

Example 6: Virtual Image of a Figurine

  • Scenario: A small porcelain figurine is positioned 5.8 cm from a converging lens, resulting in an upright virtual image with magnification 2.6.

  • Analysis: Identify the image location based on the given magnification and object distance.

Example 7: Magnification of a Picture Frame

  • Scenario: A picture frame of height 12.6 cm in front of a diverging lens forms an upright virtual image of height 4.3 cm.

  • Analysis: Calculate the magnification using:
    M=rach<em>ih</em>o=rac4.312.6M = rac{h<em>i}{h</em>o} = rac{4.3}{12.6}

Homework

  • Complete exercise set from section 13.4 on page 566: #1 - 4, 6, 8