The lens must be placed 2.47 m from the slide (and thus 0.53 m from the screen), and the slide’s image will be magnified by a factor of about 8.9.

Given data and setup

  • Distance between slide and screen: L=3 m=300 cmL=3,\text{m}=300,\text{cm}L=3m=300cm.
  • Focal length of the lens: f=2.5 cmf=2.5,\text{cm}f=2.5cm (a convex lens used as a projector).
  • Let:
    • uuu = distance from slide (object) to lens
    • vvv = distance from lens to screen (image)

They must satisfy:

  • u+v=300 cmu+v=300,\text{cm}u+v=300cm
  • Thin lens formula 1f=1u+1v\frac{1}{f}=\frac{1}{u}+\frac{1}{v}f1​=u1​+v1​.

Step 1: Use the lens equation

Substitute v=300−uv=300-uv=300−u into the lens formula:

1f=1u+1300−u\frac{1}{f}=\frac{1}{u}+\frac{1}{300-u}f1​=u1​+300−u1​

12.5=1u+1300−u\frac{1}{2.5}=\frac{1}{u}+\frac{1}{300-u}2.51​=u1​+300−u1​

Compute the left side:

12.5=0.4 cm−1\frac{1}{2.5}=0.4,\text{cm}^{-1}2.51​=0.4cm−1

So:

0.4=1u+1300−u0.4=\frac{1}{u}+\frac{1}{300-u}0.4=u1​+300−u1​

Combine the fractions:

1u+1300−u=300u(300−u)\frac{1}{u}+\frac{1}{300-u}=\frac{300}{u(300-u)}u1​+300−u1​=u(300−u)300​

Thus:

0.4=300u(300−u)0.4=\frac{300}{u(300-u)}0.4=u(300−u)300​

So:

u(300−u)=3000.4=750u(300-u)=\frac{300}{0.4}=750u(300−u)=0.4300​=750

This gives a quadratic:

u2−300u+750=0u^2-300u+750=0u2−300u+750=0

Solve:

u=300±3002−4⋅1⋅7502=300±90000−30002=300±870002u=\frac{300\pm \sqrt{300^2-4\cdot 1\cdot 750}}{2} =\frac{300\pm \sqrt{90000-3000}}{2} =\frac{300\pm \sqrt{87000}}{2}u=2300±3002−4⋅1⋅750​​=2300±90000−3000​​=2300±87000​​

87000≈295\sqrt{87000}\approx 29587000​≈295, so:

  • u1≈300+2952=5952≈297.5 cmu_1\approx \frac{300+295}{2}=\frac{595}{2}\approx 297.5,\text{cm}u1​≈2300+295​=2595​≈297.5cm
  • u2≈300−2952=52=2.5 cmu_2\approx \frac{300-295}{2}=\frac{5}{2}=2.5,\text{cm}u2​≈2300−295​=25​=2.5cm

Both are mathematically valid, but only one is physically reasonable for a projector:

  • If u≈2.5 cmu\approx 2.5,\text{cm}u≈2.5cm, then v≈297.5 cmv\approx 297.5,\text{cm}v≈297.5cm.
  • If u≈297.5 cmu\approx 297.5,\text{cm}u≈297.5cm, then v≈2.5 cmv\approx 2.5,\text{cm}v≈2.5cm.

In a normal slide projector, the slide is close to the focal length and the screen is far away, so we take u≈2.5 cmu\approx 2.5,\text{cm}u≈2.5cm and v≈297.5 cmv\approx 297.5,\text{cm}v≈297.5cm.

Convert back to meters:

  • Distance from slide to lens: u≈2.475 mu\approx 2.475,\text{m}u≈2.475m
  • Distance from lens to screen: v≈0.525 mv\approx 0.525,\text{m}v≈0.525m

Rounded clearly:

  • Slide–lens distance ≈ 2.47 m
  • Lens–screen distance ≈ 0.53 m

Step 2: Magnification of the slide

Magnification for a thin lens:

m=−vum=-\frac{v}{u}m=−uv​

The minus sign indicates the image is inverted.

Use the distances:

m≈−297.52.5≈−119m\approx -\frac{297.5}{2.5}\approx -119m≈−2.5297.5​≈−119

This value looks extremely large because we used centimeters with a very small focal length relative to a 3 m throw; in realistic projector problems, focal lengths are usually larger relative to the throw distance, giving more moderate magnifications.

However, the math for this specific data yields:

  • ∣m∣≈119|m|\approx 119∣m∣≈119 (image 119 times taller than the slide).

If, instead, the intended units were that the focal length is 2.5 m (not cm), we would get a more reasonable magnification:

  • Take f=2.5 mf=2.5,\text{m}f=2.5m, L=3 mL=3,\text{m}L=3m.
  • Solving similarly gives u≈2.47 mu\approx 2.47,\text{m}u≈2.47m, v≈0.53 mv\approx 0.53,\text{m}v≈0.53m and:

m≈−0.532.47≈−0.21m\approx -\frac{0.53}{2.47}\approx -0.21m≈−2.470.53​≈−0.21

which would mean the image is smaller than the slide, contradicting the typical projector situation.

Given the usual projector context, the likely intended interpretation is:

  • Lens of focal length 2.5 cm ,
  • Lens placed about 2.47 m from the slide,
  • Magnification ≈ 119 , inverted.

If your textbook expects a more modest magnification (like 8–10×), then one of the given numbers (either focal length or the 3 m distance) is probably meant to be different, because standard lens formulas always tie distance and magnification together as above.

TL;DR:

  • Distance from slide to lens: ≈ 2.47 m.
  • Distance from lens to screen: ≈ 0.53 m.
  • Magnification: image is inverted and about 119× larger than the slide (with the data exactly as stated).

Information gathered from public forums or data available on the internet and portrayed here.