Converging lens 15 cm focal length, object 53 cm away

[ajc9387]

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?

 

[geosonel]

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?

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

$$\frac{1}{p} \ + \ \frac{1}{q} \ = \ \frac{1}{f}$$

 

[Jameson]

In case those variables were confusing, I rewrote them.

$$\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}$$

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