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 ${~cm}$
(a) By what amount did the object change its distance from the lens? (b) What is the new magnification?
A) (a) 0.08 cm (b) 3.33
B) (a) 0.03 cm (b) 0.30
C) (a) 990.00 cm (b) 0.23
D) (a) 12.22 cm (b) 4.33

Question

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 ${~cm}$ 
(a) By what amount did the object change its distance from the lens? (b) What is the new magnification?
A) (a) 0.08 cm (b) 3.33
B) (a) 0.03 cm (b) 0.30
C) (a) 990.00 cm (b) 0.23
D) (a) 12.22 cm (b) 4.33

Quizplus · Accepted Answer

(a) Using the thin lens formula, $\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$, we can find the initial distance of the object from the lens:

$$\frac{1}{3.00}=\frac{1}{p}+\frac{1}{14.00}$$
$$p \approx 3.27 \mathrm{~cm}$$

When the new distance between the lens and sensor is 13.00 ${~cm}$, we can find the new distance between the lens and object:

$$\frac{1}{3.00}=\frac{1}{p'}+\frac{1}{13.00}$$
$$p' \approx 3.35 \mathrm{~cm}$$

The change in distance is therefore:

$$\Delta p \approx p'-p \approx 0.08 \mathrm{~cm}$$

(b) The magnification of the object is given by $\frac{q}{p}$, where $q$ is the image distance from the lens. Using the thin lens formula, we can find the initial image distance:

$$\frac{1}{3.00}=\frac{1}{14.00}+\frac{1}{q}$$
$$q \approx 3.54 \mathrm{~cm}$$

And the new image distance:

$$\frac{1}{3.00}=\frac{1}{13.00}+\frac{1}{q'}$$
$$q' \approx 3.67 \mathrm{~cm}$$

The new magnification is therefore:

$$m'=\frac{q'}{p'} \approx 3.33$$

Therefore, the best choice is A, with values (a) 0.08 cm and (b) 3.33.

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 $f _ { 0 } = 3.00 \mathrm {~cm}$