 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/section-b/9/c/solution

Use positive object distance

u = 20 cm

f = 4 cm

\frac{1}{v} = \frac{1}{4} - \frac{1}{20} = \frac{1}{5} c m^{- 1}, v = 5.0 cm .

M = - v / u = - 1/4

, so the signed height is

h_{i} = M h_{o} = - \frac{1}{4} (2 cm) = - 0.50 cm .

The real image lies $5.0 cm$ beyond the lens, is inverted, and has positive height magnitude $0.50 cm$ .

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