Answer :
Final answer:
The two converging lenses with given focal lengths form a real image with a magnification less than 1. By calculating the image distance for each lens and their respective magnifications, it's determined that the final image is real and smaller than the object. Thus, the correct answer is (b) Real, magnification < 1.
Explanation:
To solve this problem, we follow these steps:
Step 1: Find the Image Formed by the First Lens:
1//f = 1/do + 1/di
where f is the focal length, do is the object distance, and
di is the image distance. For the first lens,
f₁ = 10.1cm,
do₁ = 39.1 cm
The lens equation becomes:
1/ 10.1 = 1/ 39.11 + 1/ di₁
Solving for di₁ gives:
di₁ = 1/ (1/ 10.1 − 1/ 39.11) ≈ 13.59cm.
Since di₁ is positive, the image formed by the first lens is real and located 13.59 cm to the right of the first lens.
Step 2: Determine the Position of the Image Relative to the Second Lens:
The distance between the two lenses is 39.1 cm, so the object distance for the second lens (do₂ ) is the distance between the two lenses minus the image distance from the first lens:
do₂ = 39.1 − 13.59 = 25.51cm.
Step 3: Find the Image Formed by the Second Lens:
Using the lens equation again for the second lens (f₂ = 15.6 cm and d₂ = 25.51 cm)
1/ 15.61 = 1/ 25.51 + 1/di₂,
Solving for di₂ gives:
di₂ ≈ 40.12cm.
Since di₂ is positive, the image formed by the second lens is real and located to the right of the second lens.
Step 4: Determine the Overall Magnification:
The magnification (m) of a lens system is the product of the magnifications of each lens (m= m₁×m₂), where mi = −di/ do. Therefore,
m₁ = −di₁/ do₁ = −13.59/ 39.1,
m₂ = −di₂/ do₂ = −40.12/ 25.51.
Calculating m₁ and m₂:
m₁ ≈ −0.347,
m₂ ≈ −1.573.
Therefore, the overall magnification m is:
m = m1₁ × m₂ ≈ (−0.347) × (−1.573) ≈ 0.546.
Since the final image is real (given by the positive distance from the second lens) and the magnification is less than 1 (but positive, indicating a reduced size), the correct choice is:
b) Real, magnification < 1
