Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# 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 - Physics

Sum

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.

Concept: Refraction at Spherical Surfaces and by Lenses - Refraction at Spherical Surfaces
Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 18 Geometrical Optics
Q 53 | Page 415