- #1
JamesL
- 33
- 0
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.
Find the focal length of the lens. Answer in units of cm.
------------
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.