Question

A slide projector needs to create a 90-cm-high image of a 2.0-cm-tall slide. The screen is 250 cm from the slide.

1. What focal length does the lens need? Note the sign convention and indicate + or –
2. How far should you place the lens from the slide?

Answer 1

Using lens equation

1/f = 1/u + 1/v

And magnification of image is given by:

M = -v/u = hi/ho

Given that

ho = height of object = 2.0 cm = 2.0*10^-2 m

hi = height of image = -90 cm = -0.9 m (since in image produced on screen by projector is inverted)

M = -v/u = -0.90/(2.0*10^-2)

v = 45*u

Now also given that

u + v = distance from slide to screen = 250 cm = 2.50 m

u + v = 2.50

u + 45*u = 2.50

u = 2.50/46 = 0.0543 m = 5.43*10^-2 m

u = Object distance = 5.43 cm (Answer of part B.)

v = 45*u = 45*5.43*10^-2

v = image distance = 2.44 m

So Using values of u and v

1/f = 1/u + 1/v

f = u*v/(u + v)

f = (5.43*10^-2)*2.44/(5.43*10^-2 + 2.44)

f = +0.05311

f = +5.31*10^-2 m

f = focal length of lens = +5.31 cm (Answer of part A)

u = object distance from slide = 5.43 cm (Answer of part B)

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