A slide projector has to project a 35 mm slide (35 mm x 35 mm) on a 2m x 2m screen at a distance of 10 m from the lens. What should be the focal length of the lens in the projector?

Solution

Given, a slide projector.

Magnification, $m = \frac{- 2 m}{35 mm} = \frac{- 2 \times 10^{3}}{35}$

$∴$ $\frac{v}{u} = \frac{- 2000}{35}$

$\Rightarrow u = \frac{- 35}{2000} v = - \frac{35}{2000} \times 10 \times 10^{3} mm$

$= - 175 mm = - 0.175 m$

which is the object distance.

Now, using the lens formula,

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

$\Rightarrow$    $\frac{1}{10} + \frac{1}{0.175} = \frac{1}{f}$

$\Rightarrow$    $\frac{1}{f} = \frac{10.175}{1.75}$

$\Rightarrow$    $f = \frac{1.75}{10.175} cm = 0.172 m .$ is the required focal length of lens in the projector.

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Wave Optics

A slide projector has to project a 35 mm slide (35 mm x 35 mm) on a 2m x 2m screen at a distance of 10 m from the lens. What should be the focal length of the lens in the projector?

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