Please explain and answer how to find the Oject distance, image distance and if
ID: 1429548 • Letter: P
Question
Please explain and answer how to find the Oject distance, image distance and if the image is upright/inverted or real/vritual.
The figure is divided up into 2 parts sorry. Where the lens is at 10cm, and it appears that the arrow for the object is at 1cm and the arrown for the image is at 41.4cm.
I tried 9.cm for the object distance, and 31.4 cm for the image distance and Upright and Virtual and it is wrong.
*****picture one is cut off at 28.4cm and then continues on n picture 2, however some of the 1st picture is over lapping with the second one.
Please help. Thanks!
Explanation / Answer
We know from the lens equation that,
1/f = 1/i + 1/o
where, f is the focal length, i is the image distance and o is the object distance.
If two of the above quantities are known than the third can be calculated using the above equation.
If the image distance calculated is negative, then it is a virtual image. If it is positive then it is a real image. If it is on the same side where the object was placed than it is upright otherwise inverted.
However in the case given in question. Object distance is o = (10 - 1) = 9 cm.
Image is formed at the point where two line are intersecting. From the figure given by you, the two lines will be itersecting somewhere around 41.4 cm. So the image distance is 41.4 - 10 = 31.4 cm from the lens in its right side.
Since the image is fomed is below (on the opposite side, where the object was actually placed) so it will be inverted.
Since the image distance is positive number its a real image.
As the image looked bigger in size, it is magnified also. Magnification can be calculated using:
m = -i/o = h'/h.
(h' is the image height and h is the object's).