Step 1: Conceptual Understanding:
This problem necessitates the application of the lens formula and the magnification formula pertinent to convex lenses. The image distance can be determined using the magnification, which subsequently allows for the calculation of the focal length via the lens formula.
Step 2: Core Formulas and Methodology:
1. Magnification: \(m = \frac{h_i}{h_o} = \frac{v}{u}\), where \(h_i\) is image height, \(h_o\) is object height, \(v\) is image distance, and \(u\) is object distance.
2. Lens Formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\), where \(f\) is focal length.
3. Sign Convention (New Cartesian): Light propagates from left to right. Distances measured against the direction of incident light are negative; distances in the direction of light are positive. Object distance \(u\) is inherently negative.
Step 3: Detailed Procedure:
Provided Data:
Object height, \(h_o = 5 \, \text{cm}\).
Image height, \(h_i = 1 \, \text{cm}\).
Object distance, \(u = -40 \, \text{cm}\) (as per sign convention).
Given that the image is smaller than the object (\(h_i<h_o\)), the convex lens forms a real and inverted image. Consequently, the image height is assigned a negative value: \(h_i = -1 \, \text{cm}\).
Part 1: Image Distance (v) Calculation.
Employing the magnification formula:\[ m = \frac{h_i}{h_o} = \frac{-1 \, \text{cm}}{5 \, \text{cm}} = -\frac{1}{5} \]Utilizing the relationship \(m = \frac{v}{u}\):\[ -\frac{1}{5} = \frac{v}{-40 \, \text{cm}} \]Solving for \(v\):\[ v = (-40 \, \text{cm}) \times \left(-\frac{1}{5}\right) = +8 \, \text{cm} \]The positive value of \(v\) indicates that a real image is formed on the side of the lens opposite to the object.
Part 2: Focal Length (f) Calculation.
Applying the lens formula:\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]Substituting the values:\[ \frac{1}{f} = \frac{1}{8} - \frac{1}{-40} = \frac{1}{8} + \frac{1}{40} \]To sum the fractions, the least common denominator is 40:\[ \frac{1}{f} = \frac{5}{40} + \frac{1}{40} = \frac{6}{40} = \frac{3}{20} \]Inverting to find \(f\):\[ f = \frac{20}{3} \, \text{cm} \approx 6.67 \, \text{cm} \]
Step 4: Conclusion:
The determined focal length of the lens is approximately 6.67 cm.