Final Exam Help: Converging Lens and Light Bulb

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SUMMARY

The discussion centers on calculating the image distance for a converging lens with a focal length of 15.0 cm, positioned 53 cm from a light bulb. The relevant formula is given as 1/p + 1/q = 1/f, where d_0 represents the object distance, d_i the image distance, and f the focal length. By substituting the known values into the formula, users can determine the appropriate placement of a screen to focus the image of the light bulb.

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  • Basic knowledge of converging lenses
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Woo, this is on my final exam and i don't know if there's a formula or something to solve it. really need help. anything is appreciated.

A converging lens with a focal length of 15.0 cm is placed 53 cm from a light bulb. Where would you place a screen to focus an image of the object? :rolleyes:
 
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ajc9387 said:
Woo, this is on my final exam and i don't know if there's a formula or something to solve it. really need help. anything is appreciated.

A converging lens with a focal length of 15.0 cm is placed 53 cm from a light bulb. Where would you place a screen to focus an image of the object? :rolleyes:
click the link below to see a drawing. do you recall how to apply the formula below from your textbook?

http://img31.echo.cx/img31/1521/img12642ah.png

[tex]\frac{1}{p} \ + \ \frac{1}{q} \ = \ \frac{1}{f}[/tex]
 
In case those variables were confusing, I rewrote them.

[tex]\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}[/tex]

d_0 = object distance
d_i = image distance
f = focal length
 

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