13.4

13.4 The Thin Lens Equation

  • There is a very useful equation that relates the focal length ($f$), the object distance ($do$), and the image distance ($di$).

  • The equation is called the thin lens equation:

    • 1d<em>o+1d</em>i=1f\frac{1}{d<em>o} + \frac{1}{d</em>i} = \frac{1}{f}

Sign Convention for Thin Lens Equation

  • To use the thin lens equation, follow this sign convention:

    • Object Distance ($d_o$): Always positive.

    • Image Distance ($d_i$):

    • Positive for real images (on the opposite side of the lens from the object).

    • Negative for virtual images (on the same side of the lens as the object).

    • Focal Length ($f$):

    • Positive for converging lenses.

    • Negative for diverging lenses.

Lens Terminology

  • Variables illustrated in Figure 1:

    • $d_o$: Distance from the object to the optical centre.

    • $d_i$: Distance from the image to the optical centre.

    • $h_o$: Height of the object.

    • $h_i$: Height of the image.

    • $f$: Focal length of the lens; distance from the optical centre to the principal focus ($F$).

  • Note: The focal length ($f$) remains the same, whether measuring to $F$ or $F'$.

Derivation of the Thin Lens Equation

  1. From similar triangles in the lens system, the relationships can be established:

    • h<em>id</em>i−f=hof\frac{h<em>i}{d</em>i - f} = \frac{h_o}{f}

    • Rearranging gives:

      • h<em>i=h</em>ofdi−fh<em>i = \frac{h</em>o f}{d_i - f}

  2. Using properties of similar triangles, we find another relation:

    • d<em>id</em>o=di−ff\frac{d<em>i}{d</em>o} = \frac{d_i - f}{f}

    • Rearranging will yield:

      • 1d<em>o=1f−1d</em>i\frac{1}{d<em>o} = \frac{1}{f} - \frac{1}{d</em>i}

    • Which leads back to the thin lens equation:

      • 1d<em>o+1d</em>i=1f\frac{1}{d<em>o} + \frac{1}{d</em>i} = \frac{1}{f}

Sample Problem 1: Converging Lens

  • A converging lens has a focal length of 17 cm, and a candle is located 48 cm from the lens.

    • Given:

    • f=17 cmf = 17 \, \text{cm}

    • do=48 cmd_o = 48 \, \text{cm}

    • Required:

    • Find did_i

    • Analysis and Solution:

    1. Use the thin lens equation:

      • 1d<em>o+1d</em>i=1f\frac{1}{d<em>o} + \frac{1}{d</em>i} = \frac{1}{f}

    2. Substitute known values:

      • 148+1di=117\frac{1}{48} + \frac{1}{d_i} = \frac{1}{17}

      • 1di=117−148\frac{1}{d_i} = \frac{1}{17} - \frac{1}{48}

      • 1di=0.038 cm−1\frac{1}{d_i} = 0.038 \, \text{cm}^{-1}

      • di≈26 cmd_i ≈ 26 \, \text{cm}

    • Statement: The image of the candle is real and will be about 26 cm from the lens, opposite the object.

Sample Problem 2: Diverging Lens

  • A diverging lens with a focal length of 29 cm forms a virtual image of a marble that is 13 cm in front of the lens.

    • Given:

    • f=−29 cmf = -29 \, \text{cm}

    • di=−13 cmd_i = -13 \, \text{cm}

    • Required: Find dod_o

    • Analysis and Solution:

    1. Use the thin lens equation:

      • 1d<em>o+1d</em>i=1f\frac{1}{d<em>o} + \frac{1}{d</em>i} = \frac{1}{f}

    2. Substitute known values:

      • 1do=1−29−1−13\frac{1}{d_o} = \frac{1}{-29} - \frac{1}{-13}

      • 1do=0.043 cm−1\frac{1}{d_o} = 0.043 \, \text{cm}^{-1}

      • do≈23 cmd_o ≈ 23 \, \text{cm}

    • Statement: The marble is located 23 cm from the lens, on the same side as the image.

The Magnification Equation

  • When comparing the size of the image with the size of the object, the magnification of the lens is determined.

  • The relationship used to obtain the magnification equation:

    • h<em>ih</em>o=d<em>id</em>o\frac{h<em>i}{h</em>o} = \frac{d<em>i}{d</em>o}

  • The magnification equation can be stated as:

    • M=h<em>ih</em>o=−d<em>id</em>oM = \frac{h<em>i}{h</em>o} = -\frac{d<em>i}{d</em>o}

Sign Convention for Magnification

  • Similar sign rules apply:

    • Object Height ($h_o$): Positive when measured upward, negative when measured downward.

    • Image Height ($h_i$): Positive when measured upward, negative when measured downward.

    • Magnification ($M$): Positive for an upright image and negative for an inverted image.

  • Magnification ($M$) is dimensionless since the units cancel out.

Summary of Key Equations

  • Thin lens equation:

    • 1d<em>o+1d</em>i=1f\frac{1}{d<em>o} + \frac{1}{d</em>i} = \frac{1}{f}

  • Magnification equation:

    • M=h<em>ih</em>o=−d<em>id</em>oM = \frac{h<em>i}{h</em>o} = -\frac{d<em>i}{d</em>o}

Sign Conventions for Lenses

Variable

Positive

Negative

Object Distance ($d_o$)

Always

Never

Image Distance ($d_i$)

Real image (opposite side)

Virtual image (same side)

Height of Object ($h_o$)

Measured upward

Measured downward

Height of Image ($h_i$)

Measured upward

Measured downward

Focal Length ($f$)

Converging lens

Diverging lens

Magnification ($M$)

Upright image

Inverted image

Practice Problems

  1. Calculate the image location for a frog with a focal length of 23 cm and distance of 32 cm from a converging lens.

  2. Determine the focal length for a pencil located 53 cm from a diverging lens with a virtual image at 18 cm.

  3. For a diverging lens with a focal length of 34 cm, locate a booklet positioned 13 cm behind the lens.

  4. Find the image of an insect located 11 cm from a converging lens with a focal length of 16 cm.

  5. Use the magnification equation to analyze a vase of height 12 cm with an inverted image of height 35 cm.

  6. Calculate the magnification for a playing card of height 14 cm, resulting in an inverted real image height of 7.9 cm.

  7. For a postage stamp of height 2.8 cm in front of a diverging lens, find the magnification of a virtual image of height 1.3 cm.

  8. Determine the location and focal length for a fork placed 9.4 cm before a lens creating an upright virtual image with a magnification of 5.6.

1. For the image location with a focal length of 23 cm and object distance of 32 cm: - Use the thin lens equation: 1do+1di=1f\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}. - Substitute values: 132+1di=123\frac{1}{32} + \frac{1}{d_i} = \frac{1}{23}. - Solve for did_i to find the image distance. 2. For the focal length of the pencil with an object distance of 53 cm and a virtual image at -18 cm: - Use the thin lens equation: 1do+1di=1f\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}. - Substitute values: 153+1−18=1f\frac{1}{53} + \frac{1}{-18} = \frac{1}{f}. - Solve for ff. 3. For locating a booklet with a diverging lens and focal length of -34 cm at -13 cm: - Use the thin lens equation: 1do+1di=1f\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}. - Substitute values and solve for dod_o. 4. For finding the image of an insect at 11 cm from a converging lens with a focal length of 16 cm: - Use the thin lens equation again: 111+1di=116\frac{1}{11} + \frac{1}{d_i} = \frac{1}{16}. - Solve for did_i. 5. To analyze the magnification for a vase of height 12 cm and an inverted image of height -35 cm: - Use the magnification equation: M=hihoM = \frac{h_i}{h_o}, plug in values for magnification calculation. 6. For the playing card of height 14 cm producing a real inverted image of height -7.9 cm: - Use M=−didoM = -\frac{d_i}{d_o} and find did_i and dod_o based on the magnification. 7. For the postage stamp with a diverging lens: - Use magnification M=heightofvirtualimageheightofobject=1.32.8M = \frac{height of virtual image}{height of object} = \frac{1.3}{2.8} to find MM. 8. For a fork producing an upright virtual image with a magnification of 5.6 at 9.4 cm: - Use M=hihoM = \frac{h_i}{h_o} and solve to find dod_o and ff.