How to find the position of the final image of the 1.0-cm-tall object in Physics? Can you find the position of the final image of the 2.0-cm-tall object?
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You will observe questions like find the position of the final image of the 1.0-cm-tall object in subjects such as Physics and science. You have to understand the question description properly first before attempting to solve the same.
We have to solve it as per the Physics formulae that we use for while calculating values of mirror or lens.
(Image Source: https://www.physicsforums.com/threads/double-lens-system-converging-diverging.805553/)
Solved: Find the position of the final image of the 1.0-cm-tall object (Using Formula)
You have to put the values as follows:
Here, we have to find the value of di first of the 1st lens.
Substitution the values of each digit in equation – (i), we get the following figure:
By solving the above equation, we get di (distance) = -20-cm (Because it is located to the left side of 1st lens)
Now, adding the distance between lenses, we have to go through the same formula for the 2nd lens as well.
Now, using the same formula, we can put values as follows:
If you calculate the equation, you will get di= -6.4-cm.
Thus, the final image distance is -6.4-cm located to the left of the 2nd lens.
When we use magnification formula, -V1/ u1 * -V2/ u2
Image Height= M X Height of the Object
It means the image so formed will be erect.
How to Find the position of the final image of the 2.0-cm-tall object.
Reference: A 2.0-cm tall object is positioned 40-cm away from a diverging lens with a focal length of 15 cm. Find the position and size of the image.
If we go by the above formula, we have to calculate it as follows:
We know that Concave Lens = Diverging Lens
Here, Focal Length= 15-Cm
The distance of the object from the lens= 40-cm
Height of the object= 2.0-cm
We have to find the position of the image along with its height of the object.
Going by the lens formula, we know:
By putting the values in above formula, we get the following:
It means the image distance from the lens will be 10.9-cm. The use of negative sign before the distance means it will be formed to the left side or behind the lens.
For further analysis, we can also use magnification formula as here:
Putting the values in their respective positions, we get;
Solving the above sum; we get
It means the height of the image is + 0.54. And the image thus obtained will be erect and virtual.
Thus, the image distance from the lens is 10.90 cm, and the negative sign implies that it is formed behind the lens (on the left side).