# Homework Help: Quick Question about lenses/refraction

1. Nov 18, 2004

### JamesL

A concave lens forms a virtual image .5 times the size of the object. The distance between the object and the image is 7.9cm.

Find the focal length of the lens. Answer in units of cm.
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i suppose what is confusing me is whether or not the object itself is virtual. i dont see how (with a concave lens) a real object could produce a virtual image smaller than itself.

i can construct a series of equations for this problem fairly easily, but im not doing it correctly apparently.

anyway, if i consider the object itself to be virtual, both Id and Od (image distance and object distance would be negative)...

so:

.5 = -(-Id)/(-Od)
Od = -2Id **

-Od - Id = 7.9 cm **

focal length f = ((1/-2Id) - (1/Id))^-1 **

the equations with asteriks were the ones i used... am i setting them up correctly?

any help is appreciated.

2. Nov 18, 2004

### Staff: Mentor

sign convention

But there's no reason to assume a virtual object. Assume the usual real object and let the equations do the work. But you'd better use a consistent sign convention: if you want Id to be a positive number, then the image distance is -Id; similarly, the object distance (assumed real until proven otherwise) is the positive number Od.

0.5 = - (-Id)/(Od)
Od = 2Id

Od - Id = 7.9 cm
Thus: Id = 7.9 cm; Od = 2(7.9) cm

Now use the lens equation to find the focal length (which better turn out negative!):
1/f = 1/Od + 1/(-Id)
... etc

(Tip: Draw yourself a picture.)