Thin Lens Equation Problems Solutions

Thin Lens Equation Problems Solutions

Thin Lens Equation Definition

• The Thin Lens Equation relates to lenses and states that:

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

  • where:

    • $f$ - focal length of the lens

    • $d_o$ - object distance

    • $d_i$ - image distance

  • Note: All measurements must be in the same units (all in cm or all in m, etc.)

Magnification Equation Definition

• The Magnification Equation describes the relationship between the heights of the object and image and their distances from the lens:

  • $M = \frac{h<em>i}{h</em>o} = -\frac{d<em>i}{d</em>o}$

  • where:

    • $M$ - magnification

    • $h_o$ - object height

    • $h_i$ - image height

    • The negative sign indicates that the image is inverted.

Example Problems

Problem 1: Converging Lens Focal Length

• Given: An object located 25 cm from the lens produces a focused image on film 15 cm away.

  • Known Values:

  • $d_o = 25 \, \text{cm}$

  • $d_i = 15 \, \text{cm}$

• Find: Focal length, $f$.

• **Calculation: *Using thin lens equation:*

  • $\frac{1}{f} = \frac{1}{d<em>o} + \frac{1}{d</em>i} = \frac{1}{25} + \frac{1}{15}$

  • Calculating gives:

  • $\frac{1}{25} = 0.04$

  • $\frac{1}{15} = 0.06667$

  • $\frac{1}{f} = 0.04 + 0.06667 = 0.10667$

  • Thus, $f = 9.375$ cm (rounded to 9.4 cm).

Problem 2: Object Distance for Camera Lens

• Given: Focal length of lens is 2 cm; film is set 3 cm from lens.

  • Known Values:

  • $d_i = 3 \, \text{cm}$

  • $f = 2 \, \text{cm}$

• Find: Object distance, $d_o$.

• Calculation:

  • Using thin lens equation:

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

  • $\frac{1}{2} = \frac{1}{d_o} + \frac{1}{3}$

  • Rearranging gives:

    • $\frac{1}{d_o} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6}$

  • Therefore, $d_o = 6 \, \text{cm}$.

Problem 3: Converging Lens and Image Size of a Tree

• Given: Tree height is 20 m, distance from lens is 40 m, focal length 8.0 cm.

  • Known Values:

  • $h_o = 20 \, \text{m} = 2000 \, \text{cm}$

  • $d_o = 40 \, \text{m} = 4000 \, \text{cm}$

  • $f = 8.0 \, \text{cm}$

• Calculations:

  1. Distance from lens to image, $d_i$: Using thin lens equation:

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

  • Rearranging yields:

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

    • $\frac{1}{d_i} = \frac{1}{8.0} - \frac{1}{4000} = 0.125 - 0.00025 = 0.12475$

  • Therefore, $d_i = 8.0 \, \text{cm}$.

  1. Calculating magnification, $M$:

  • Using magnification equation:

  • $M = -\frac{d<em>i}{d</em>o} = -\frac{8.0}{4000} = -0.0020$

  1. Height of the image of the tree, $h_i$:

  • $M = \frac{h<em>i}{h</em>o}$

  • $-0.0020 = \frac{h_i}{2000}$

  • Thus, $h_i = -4 \, \text{cm}$ (negative indicates it is inverted).

Problem 4: Image Distance in Human Eye

• Given: Focal length of the human eye is 2.3 cm, looking at an object 55.3 cm away.

• Find: Image distance, $d_i$.

• Calculation:

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

  • $\frac{1}{2.3} = \frac{1}{55.3} + \frac{1}{d_i}$

  • Rearranging yields:

  • $\frac{1}{d_i} = \frac{1}{2.3} - \frac{1}{55.3}$

  • $\approx 0.43478 - 0.0181\approx 0.41768$

  • Finally, $d_i = 2.40 \text{cm}$.

Problem 5: Converging Lens Size Relations

• Given: Converging lens with an object positioned 40 cm away and produces an image twice the size.

• Calculations:

  1. Image Distance, $d_i$: Using magnification

  • $M = -2 = -\frac{d_i}{40}$

  • Thus, $d_i = 80 \, \text{cm}$.

  1. Focal Length, $f$:

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

  • $\frac{1}{f} = \frac{1}{40} + \frac{1}{80} = \frac{2}{80} + \frac{1}{80} = \frac{3}{80}$

  • Therefore, $f = 26.7 \text{cm}$.

  1. Original Object Height, $h<em>o$: Given that image height $h</em>i = 6 cm$,

  • $\frac{h<em>i}{h</em>o} = -2$

  • Thus, $h<em>o = \frac{h</em>i}{-2} = \frac{6}{2} = 3 \text{cm}$.

Problem 6: Eye Lens Focal Length

• Given: Focal length of lens in the eye is 1.5 cm; maximum image size perceived is 5 cm; object positioned 100 cm away.

• Calculations:

  1. Image Distance, $d_i$:

  • Using thin lens equation:

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

    • Thus:

      • $\frac{1}{1.5} = \frac{1}{100} + \frac{1}{d_i}$

    • Rearranging gives:

      • $\frac{1}{d_i} = \frac{1}{1.5} - \frac{1}{100} = 0.6667 - 0.01 = 0.657$

    • Therefore, $d_i = 1.52 \text{cm}$.

  1. Maximum Size to See the Object:

  • Using magnification,

  • $M = \frac{h<em>i}{h</em>o} = -\frac{d<em>i}{d</em>o}$

    • Thus, using $d<em>i = 1.52 \,\text{cm}$ and $d</em>o = 100 \, \text{cm}$:

    • $-0.0152 = -\frac{h_i}{100}$

    • Rearranging yields:

      • $h_i = 0.0152 \times 100 = 1.52 \text{cm}$

    • If $M = 5, ext{ (maximum height) }$

      • Hence, max size: $5 \text{cm} = -329 \text{cm}$ for complete view.

Conclusion

• Understanding the Thin Lens Equation and its practical applications is crucial in optics, especially in the field of photography and vision science.

• The calculations for focal lengths, distances, and heights provide essential insights into the behavior of lenses, whether in cameras, microscopes, or the human eye, highlighting the significance of real and virtual images.