A simple digital camera lens with focal length $f _ { 0 } = 3.00 \mathrm {~cm}$ perfectly focuses the image of an object on a CCD sensor located a distance $14.00 \mathrm {~cm}$ away from the lens.The object now moves away from the lens and the new distance between the lens and sensor is $13.00 \mathrm {~cm}$ (a)By what amount did the object change its distance from the lens? (b)What is the new magnification?

Question

Quizplus · Accepted Answer

Using the thin lens formula, $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$, we can find the initial object distance:
$\frac{1}{3}=\frac{1}{d_o}+\frac{1}{14}$
Solving for $d_o$, we get $d_o=4.667 \mathrm{~cm}$
Using the same formula, we can find the new object distance:
$\frac{1}{3}=\frac{1}{d_o'}+\frac{1}{13}$
Solving for $d_o'$, we get $d_o'=4.750 \mathrm{~cm}$
Therefore, the object moved away by $4.750-4.667=0.083 \mathrm{~cm}$, or 0.08 cm (rounded to 2 significant figures).
To find the new magnification, we can use the formula $\frac{m'}{m}=\frac{d_i'}{d_i}=\frac{d_o}{d_o'}$. Plugging in the values we found, we get:
$\frac{m'}{m}=\frac{4.667}{4.750}\approx0.966$
Therefore, the new magnification is $m'=0.966\times m$.
Using the thin lens formula again, we can find the magnification using the initial object distance:
$\frac{1}{3}=\frac{1}{d_i}-\frac{1}{d_o}$
Solving for $d_i$, we get $d_i=12 \mathrm{~cm}$, so the initial magnification is $\frac{12}{3}=4$.
Therefore, the new magnification is approximately $4\times 0.966$, which is 3.33 (rounded to 2 significant figures).

A Simple Digital Camera Lens with Focal Length f0=3.00 cmf _ { 0 } = 3.00 \mathrm {~cm}f0​=3.00 cm

A Simple Digital Camera Lens with Focal Length $f _ { 0 } = 3.00 \mathrm {~cm}$