## Related questions with answers

An f/2.80 CCD camera has a 105-mm focal length lens and can focus on objects from infinity to as near as 30.0 cm from the lens. maximum distances from the CCD sensor over which the lens must be able to travel during focusing. Note: ''f/2.80" means "an f-number of 2.80."

Solution

VerifiedWhen the object is 30 cm away from the camera ($x_{min}$), the image is created at the distance of $x_2$ from the lens.

$\frac{1}{x_{min}} + \frac{1}{x_2} = \frac{1}{f}$

Solving this for $x_2$:

$x_2 = \frac{1}{ \frac{1}{f} - \frac{1}{x_{min}} } = \frac{1}{ \frac{1}{10.5 \text{ cm}} - \frac{1}{30 \text{ cm}} } = \boxed{ 16.154 \text{ cm}}$

The distance of the image from the lense is also the distance of the lense from the CCD sensor since we want the image to be projected on it.

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