Question

A camera with a 90-mm-focal-length lens is focused on an object 1.30 m from the lens. To refocus on an object 6.50 m from the lens, by how much must the distance between the lens and the sensor be changed

Answer 1

Answer:

The distance between the lens and the sensor must be changed by 5.454-mm

Explanation:

$\frac{1}{S} +\frac{1}{S^'}} = \frac{1}{f}$

Given;

the focal length = 90-mm

object distance = 1.30 m= 1300-mm

$\frac{1}{S^'}} = \frac{1}{f} - \frac{1}{S} \\\\\frac{1}{S^'}} = \frac{1}{90} - \frac{1}{1300}\\\\\frac{1}{S^'}} = \frac{121}{11700} \\\\S{^'} = \frac{11700}{121} =96.694-mm$

When the object is refocused from the lens;

New object distance = 6.50 m = 6500-mm

$\frac{1}{S^'}} = \frac{1}{f} - \frac{1}{S} \\\\\frac{1}{S^'}} = \frac{1}{90} - \frac{1}{6500}\\\\\frac{1}{S^'}} = \frac{641}{58500} \\\\S{^'} = \frac{58500}{641} =91.264-mm$

determine how much the distance between the lens and the sensor be changed; = Δs'

Δs' = (96.694 - 91.264)mm = 5.454-mm

Therefore, the distance between the lens and the sensor must be changed by 5.454-mm