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

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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
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APPEARS IN

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