 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/section-a/3/solution

u, v

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}, v = \frac{f u}{u - f} = \frac{( 4 ) ( 300 )}{300 - 4} mm = 4.054 mm .

The transverse magnification is

M = - v / u = - f / (u - f) = - 4/296

. For an object dimension

500 mm

, the corresponding image dimension is

∣ h_{i} ∣ = \frac{4}{296} (500) mm = 6.76 mm .

The minus sign of the magnification means that the image is inverted. This is the ideal thin-lens prediction; it assumes the relevant field is admitted by the camera and ignores lens distortion.

 New to topics? Read the docs here!