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Topic Content:
- Lens Formula
- Sign Convention
- Magnification
- Worked Examples
Findings show that there is a relationship between object distance ( u ), image distance ( v ), and focal length ( f ), of a lens. This relationship is given by:
\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)( Real is Positive )
\(\frac{1}{f} = \frac{1}{v} \: – \: \frac{1}{u} \)( New Cartesian )
Sign Convention:
For the sign conventions:
Tip: Try and master one sign convention and stick with it for all your calculations. Our examples will focus on the New Cartesian Sign Convention.
Magnification:
For both convex and concave lenses, the magnification produced is given by:
m = \( \normalsize \frac{Image \; distance}{object \; distance} = \frac{v}{u}\)
m = \(\normalsize \frac{Image \; height}{object \; height} = \frac{h_i}{h_o} \)
m = \( \frac{v}{f} \scriptsize \; – \; 1 \)( Real is positive )
m = \( \: – \frac{v}{f} \scriptsize \; + \; 1 \)( New Cartesian )
Worked Examples:
7.5.1. If an object of 5 cm height is placed at a distance of 15 cm from a convex lens of focal length 10 cm, find the position, nature and height of the image.
7.5.2. A convex lens of focal length 12 cm forms a real image 36 cm from the lens. What is the magnification and size of the image if the object height is 3 cm?
7.5.3. A lens of focal length 12.0
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