Two converging lenses, having focal length 5 cm and 10 cm respectively, are placed
25cm apart. An object is placed 10 cm in front of the lens of focal length 5cm.
Determine the position and magnification of the final image formed.
To determine the position and magnification of the final image formed by the two converging lenses, we need to calculate the intermediate image formed by the first lens and then use it as the object for the second lens.
Let's start by calculating the position and magnification of the image formed by the first lens.
Given:
Focal length of the first lens (f1) = 5 cm
Object distance from the first lens (p1) = 10 cm
We can use the lens formula to calculate the image distance from the first lens (q1) and the magnification (m1).
The lens formula is given by:
1/f1 = 1/p1 + 1/q1
Substituting the given values into the formula:
1/5 = 1/10 + 1/q1
Simplifying the equation:
1/q1 = 1/5 - 1/10
1/q1 = (2 - 1)/10
1/q1 = 1/10
q1 = 10 cm
To calculate the magnification (m1), we can use the magnification formula:
m1 = -q1/p1
Substituting the values:
m1 = -10/10
m1 = -1
The negative sign indicates an inverted image formed by the first lens.
Now we have the image formed by the first lens. We can consider this image as the object for the second lens.
Given:
Focal length of the second lens (f2) = 10 cm
Object distance from the second lens (p2) = 10 cm (the image distance formed by the first lens)
To calculate the image distance from the second lens (q2) and the magnification (m2), we can use the lens formula, similar to the first lens.
The lens formula for the second lens becomes:
1/f2 = 1/p2 + 1/q2
Substituting the given values:
1/10 = 1/10 + 1/q2
Simplifying the equation:
1/q2 = 1/10 - 1/10
1/q2 = 0
q2 = infinity
From the above calculation, we can see that the image distance (q2) from the second lens is infinity. This indicates that the final image formed is at infinity.
To calculate the magnification (m2), we use the magnification formula:
m2 = -q2/p2
Substituting the values:
m2 = - ∞ / 10
m2 = -∞
Therefore, the position of the final image formed by the two converging lenses is at infinity, and the magnification is negative infinity (or simply, undefined).
So, the final image formed is an inverted virtual image at infinity, and it is highly magnified compared to the size of the original object.