Concave lens with virtual image 0.5 times object size

[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 don't 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 I am 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.

 

[Doc Al]

sign convention

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

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.

so:

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

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

-Od - Id = 7.9 cm **

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.)

 

[wais]

Hi there, it seems like you are on the right track with your equations. To find the focal length of the lens, we can use the thin lens equation: 1/f = 1/di + 1/do, where f is the focal length, di is the image distance, and do is the object distance.

In this case, we know that the image distance is 7.9 cm and the object distance is negative since it is virtual. So, we can plug in these values to get:

1/f = 1/7.9 + 1/-Od

Now, we also know that the image size is 0.5 times the object size, so we can use the magnification equation: m = -di/do = hi/ho, where m is the magnification, hi is the image height, and ho is the object height.

In this case, since the image is virtual, we can say that hi is negative and ho is positive. So, we can write:

-0.5 = -7.9/Od

Solving for Od, we get Od = 15.8 cm.

Now, going back to our thin lens equation, we can plug in this value for Od and solve for f:

1/f = 1/7.9 + 1/-15.8

Solving for f, we get f = -11.85 cm.

Since the focal length is a distance, we can say that the focal length is 11.85 cm.

Hope this helps clarify things for you! Let me know if you have any other questions. Happy to help!