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

Answer: (A)

The final image will be at f/2 behind the rightmost lens and it's magnification will be -1

Answer 2

When three identical convex lenses are positioned in a straight line, each with a focal length of f , separated by a distance f from each other, and an object is placed at 2=2 f ​=f in front of the leftmost lens, a specific image formation occurs.

Due to the arrangement, the image formed by the first lens acts as the object for the second lens, and similarly, the image formed by the second lens serves as the object for the third lens. This creates a setup similar to a single lens.

In such an arrangement, the final image will be formed behind the rightmost lens at a distance of 2 f ​ from it. This is because the combined focal lengths and distances create an effective focal length for the entire system.

Additionally, the magnification of this system will be -1, which indicates that the image is inverted. This occurs because the object is placed at the focal point of the first lens, creating an image that acts as a real and inverted object for the next lens.

In summary, due to the arrangement of the identical convex lenses, the image will be located 2 f ​ behind the rightmost lens with a magnification of -1. This outcome aligns with the properties of lens systems and the formation of images.

The correct option is(A): The final image will be at f/2 behind the rightmost lens and it's magnification will be -1