Question

Three identical convex lenses each of focal length f are placed in a straight line separated by a distance f from each other. An object is located in \(\frac{f}{2}\) in front of the leftmost lens. Then,

• A. The final image will be at f/2 behind the rightmost lens and it's magnification will be -1
• B. The final image will be at f/2 behind the rightmost lens and it's magnification will be -1
• C. The final image will be at f behind the rightmost lens and it's magnification will be -1
• D. The final image will be at f behind the rightmost lens and it's magnification will be +1

Answer 1

The Correct Option is A

When three identical convex lenses are placed in a line, each with a focal length of $f$ and separated by a distance $f$, and an object is placed at a distance $2f$ from the first (leftmost) lens, we analyze the image formation as follows:

1. First Lens: 
The object is at $2f$, so it forms a real, inverted image at a distance $2f$ on the other side of the first lens.

2. Second Lens:
The image from the first lens is now at a distance $f$ in front of the second lens (since lenses are $f$ apart). This means it is at the focal point of the second lens, producing an image at infinity.

3. Third Lens:
The rays coming from the second lens are parallel (since image was at infinity), and when these parallel rays pass through the third lens, they converge at the focal point of the third lens — that is, at a distance $f$ behind the third lens.

Final Image:
The final image is formed at a distance $f$ behind the third (rightmost) lens.

Magnification:
The magnification by the first lens is $-1$ (since object at $2f$ gives image at $2f$, inverted). The second lens sends rays to infinity (magnification undefined), but the third lens brings parallel rays to a point — thus overall magnification remains $-1$.

Correct Answer: The final image is at a distance $f$ behind the rightmost lens and the magnification is $-1$.