A slide projector is required to project a 35 mm slide ( 35 mm × 23...

To project the entire slide onto the screen, the maximum magnification is determined by fitting the larger dimension of the slide ($35 \text{ mm}$) onto the screen ($2 \text{ m}$).

Magnification $m = \dfrac{v}{|u|} = \dfrac{2000 \text{ mm}}{35 \text{ mm}} = \dfrac{400}{7}$

Given the image distance $v = 10 \text{ m} = 1000 \text{ cm}$, the object distance $|u|$ is:

$|u| = \dfrac{v}{m} = \dfrac{1000}{\dfrac{400}{7}} = 17.5 \text{ cm}$

Using the sign convention, $u = -17.5 \text{ cm}$ and $v = +1000 \text{ cm}$. Applying the lens formula:

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$\dfrac{1}{f} = \dfrac{1}{1000} -$\left(-\dfrac{1}{17.5}\right)

\dfrac{1}{f} = \dfrac{1}{1000} + \dfrac{1}{17.5} = \dfrac{17.5 + 1000}{17500} = \dfrac{1017.5}{17500}

$f =$\dfrac{17500}{1017.5} \approx$17.2 \text{ cm}$