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A camera is equipped with a lens with a focal length of \( 32 \mathrm{~cm} \). When an object \( 2.6 \mathrm{~m}(260 \mathrm{~cm}) \) away is being photographed, what is the magnification? \[ 2 \]

User Rana Depto
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To solve this problem, we are going to use two composed formulas, one called the Lens Equation and the other called Magnification formula.

The Lens equation is:

1 / f = 1 / v - 1 / u.

Here, f is the focal length of the lens, u is the object distance and v is the image distance. These measures are commonly set in centimeters, but any given unit can be used as long it's consistent with all the other measures. In our case, the focal length is given as 32 cm, and the object distance is given as 260 cm.

Rewriting the equation to solve for image distance (v), we will have the equation as:

v = 1 / [1 / f + 1 / u].

By substituting the focal length f = 32 cm and the object distance u = 260 cm into the formula, the image distance v is calculated as 28.493150684931507 cm.

The second part of the question asks for the magnification. Magnification is defined as the image distance divided by the object distance (m = v / u). The value of this equation for our data is 28.493150684931507 cm / 260 cm = 0.1095890410958904.

So, for a lens with a focal length of 32 cm photographing an object 260 cm away, the image distance will be approximately 28.49 cm and the magnification will be approximately 0.11.

User Zynk
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