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An object is placed 30.0 cm from a converging lens. If the real image formed is 90.0 cm from the object, what is the focal length?

Sagot :

To find the focal length of a converging lens when an object is placed at a certain distance from the lens and a real image is formed at another distance, we use the lens formula:
[tex]\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \][/tex]
where:
- [tex]\( f \)[/tex] is the focal length of the lens,
- [tex]\( v \)[/tex] is the distance of the image from the lens,
- [tex]\( u \)[/tex] is the distance of the object from the lens.

Given:
- The object distance [tex]\( u \)[/tex] is 30.0 cm,
- The image distance [tex]\( v \)[/tex] is 90.0 cm.

Step-by-step solution:

1. Convert the given distances into their respective reciprocals for the lens formula.
[tex]\[ \frac{1}{u} = \frac{1}{30.0} \][/tex]
[tex]\[ \frac{1}{v} = \frac{1}{90.0} \][/tex]

2. Add these reciprocals according to the lens formula.
[tex]\[ \frac{1}{f} = \frac{1}{90.0} + \frac{1}{30.0} \][/tex]
3. Now, compute:
[tex]\[ \frac{1}{f} = \frac{1}{90.0} + \frac{1}{30.0} = 0.011111111111111112 + 0.03333333333333333 \][/tex]
[tex]\[ \frac{1}{f} = 0.044444444444444446 \][/tex]

4. To find the focal length [tex]\( f \)[/tex], take the reciprocal of the sum.
[tex]\[ f = \frac{1}{0.044444444444444446} \][/tex]

5. Calculate the reciprocal:
[tex]\[ f = 22.5 \text{ cm} \][/tex]

Hence, the focal length of the converging lens is [tex]\( 22.5 \)[/tex] cm.
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