A student was asked to measure the focal length of a converging lens - Leaving Cert Physics - Question 2 - 2009

1. Measure from the center of the lens: Ensure that measurements are taken from the optical center of the lens to the screen to avoid inaccuracies.
2. Avoid Parallax Errors: Ensure that the measuring device is perpendicular to the path of light and the screen to prevent parallax errors when reading measurements.

To calculate the focal length (f) of the converging lens, we use the lens formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

Using the provided data:

• For u = 20.0 cm, v = 65.2 cm: $\frac{1}{f} = \frac{1}{20.0} + \frac{1}{65.2}$
• For u = 30.0 cm, v = 33.3 cm: $\frac{1}{f} = \frac{1}{30.0} + \frac{1}{33.3}$
• For u = 40.0 cm, v = 25.1 cm: $\frac{1}{f} = \frac{1}{40.0} + \frac{1}{25.1}$

Calculating these:

1. First data: $\frac{1}{f} = 0.05 + 0.0153 \approx 0.0653 \Rightarrow f \approx 15.3 \text{ cm}$
2. Second data: $\frac{1}{f} = 0.0333 + 0.0300 \approx 0.0633 \Rightarrow f \approx 15.8 \text{ cm}$
3. Third data: $\frac{1}{f} = 0.025 + 0.0398 \approx 0.0648 \Rightarrow f \approx 15.4 \text{ cm}$

After using all data: The average focal length is approximately $f = 15.5 \pm 0.4 \text{ cm}$.

If the student placed the object 10 cm from the lens, they would encounter the following difficulties:

1. Object Inside Focal Length: The object is closer than the focal length of the lens, leading to a virtual image that cannot be projected onto a screen.
2. Difficulty Locating Image: The image formed would be virtual, making it challenging to locate accurately without a parallax method.