A slide projector has to project a 35 mm slide (35 mm × 23 mm) on a 2 m × 2 m screen at a distance of 10 m from the lens. What should be the focal length of the lens in the projector?

#### Solution

Given,

We are projecting a slide of 35 mm \[\times\] 23 mm on a 2 m

`xx` 2 m screen using projector.

Therefore, the magnification required by the projector is: `m= v/u`

here,*v* = Image distance*u* = Object distance

We will take 35 mm as the object size

∵ 35 mm > 23 mm

\[\therefore m = \frac{v}{u} = \frac{h_i}{h_o} \]

\[\frac{v}{u} = \frac{2 \times {10}^3}{35}\]

\[ \Rightarrow u = \frac{35}{2 \times {10}^3} \times v\]

\[ \Rightarrow u = \frac{35}{2 \times {10}^3} \times 10\]

\[ \Rightarrow u = 0 . 175 \text{ mm }\]

The lens formula is given by

\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]

\[\Rightarrow \frac{1}{10} + \frac{1}{0 . 175} = \frac{1}{f}\]

\[\Rightarrow \frac{10175}{1750} = \frac{1}{f}\]

\[\Rightarrow f = \frac{1750}{10175}\]

\[ f = 0 . 172 \text{ mm }\]

Hence, the required focal length is 0.172 mm.