The problem is about focusing an image of a slide located 3 meters from a screen using a lens with a focal length of 2 cm. The goal is to find the distance from the slide at which the lens should be placed to form a focused image on the screen and to calculate the magnification of the slide. We can use the lens formula for this:

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}f1​=v1​−u1​

where:

• fff is the focal length of the lens (2 cm in this case),
• uuu is the object distance from the lens (distance of the slide to the lens),
• vvv is the image distance from the lens (distance of the lens to the screen).

Also, the distance between the slide and the screen is fixed:

u+v=300 cmu+v=300\text{ cm}u+v=300 cm

We will solve these equations to find uuu and vvv, and then determine the magnification:

M=vuM=\frac{v}{u}M=uv​

I'll calculate the values now.Let's denote:

• uuu = distance from slide to lens (to be found),
• vvv = distance from lens to screen = 300−u300-u300−u cm,
• f=2f=2f=2 cm (focal length).

The lens formula is:

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}f1​=v1​−u1​

Substitute v=300−uv=300-uv=300−u:

12=1300−u−1u\frac{1}{2}=\frac{1}{300-u}-\frac{1}{u}21​=300−u1​−u1​

Multiply through by 2u(300−u)2u(300-u)2u(300−u):

u(300−u)=2u−2(300−u)u(300-u)=2u-2(300-u)u(300−u)=2u−2(300−u)

Rewrite:

u(300−u)=2u−600+2uu(300-u)=2u-600+2uu(300−u)=2u−600+2u

300u−u2=4u−600300u-u^2=4u-600300u−u2=4u−600

Rearranged:

u2−296u−600=0u^2-296u-600=0u2−296u−600=0

Now solve this quadratic equation for uuu:

u=296±2962+4×6002u=\frac{296\pm \sqrt{296^2+4\times 600}}{2}u=2296±2962+4×600​​

Calculate uuu and then calculate magnification M=vu=300−uuM=\frac{v}{u}=\frac{300-u}{u}M=uv​=u300−u​. Calculating these values now.Using the quadratic formula:

u=296±2962+4×6002u=\frac{296\pm \sqrt{296^2+4\times 600}}{2}u=2296±2962+4×600​​

Calculate the discriminant:

2962=87616,4×600=2400296^2=87616,\quad 4\times 600=24002962=87616,4×600=2400

87616+2400=9001687616+2400=9001687616+2400=90016

90016≈300\sqrt{90016}\approx 30090016​≈300

Thus,

u=296±3002u=\frac{296\pm 300}{2}u=2296±300​

Two possible solutions:

1. u=296+3002=5962=298u=\frac{296+300}{2}=\frac{596}{2}=298u=2296+300​=2596​=298 cm
2. u=296−3002=−42=−2u=\frac{296-300}{2}=\frac{-4}{2}=-2u=2296−300​=2−4​=−2 cm (not physically valid for object distance)

So, u=298u=298u=298 cm. Then,

v=300−u=300−298=2 cmv=300-u=300-298=2,\text{cm}v=300−u=300−298=2cm

Magnification:

M=vu=2298≈0.0067M=\frac{v}{u}=\frac{2}{298}\approx 0.0067M=uv​=2982​≈0.0067

So, the lens must be placed approximately 298 cm (about 3 meters) from the slide, and the image will be a very small magnification of about 0.0067. This suggests the lens is very close to the screen (2 cm), near the focal length. Answer:

• The lens should be placed about 298 cm from the slide.
• The slide will be magnified approximately 0.0067 times (very small image).

If a bigger magnification is desired, the lens could be placed differently but the task was to focus the image on the screen 3 meters away. These calculations are consistent with lens optics principles using the thin lens equation.