An object is placed 10 cm in front of a lens. The image formed is real, inverted and of same size as the object. What is the focal length and nature of the lens?
5 cm, converging
10 cm, diverging
20 cm, converging
20 cm, diverging
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CAPF – 2021
The image formed is real, inverted, and of the same size as the object.
A real and inverted image is formed by a converging lens (convex lens). Diverging lenses (concave lenses) always produce virtual, erect, and diminished images.
For a lens, the magnification (m) is given by the ratio of image distance (v) to object distance (u), with a sign convention: m = v/u for erect images and m = -v/u for inverted images.
Since the image is inverted, m = -v/u.
The image is of the same size as the object, so the magnitude of magnification |m| = 1.
Thus, m = -1.
-v/u = -1 => v = u.
Given u = 10 cm, so v = 10 cm.
Using the lens formula (thin lens equation): 1/f = 1/v – 1/u.
Applying the sign convention: object distance u is typically taken as negative when placed in front of the lens, so u = -10 cm. The image is real and formed on the opposite side of the lens from the object, so image distance v is positive, v = +10 cm.
1/f = 1/(+10 cm) – 1/(-10 cm)
1/f = 1/10 + 1/10
1/f = 2/10
1/f = 1/5
f = 5 cm.
The focal length is positive (f > 0), which confirms that the lens is a converging lens (convex lens).
Alternatively, for a converging lens, a real image of the same size as the object is formed only when the object is placed at a distance of 2f from the lens, and the image is also formed at 2f on the other side. So, u = 2f.
Given u = 10 cm, 10 cm = 2f => f = 5 cm.
The nature of the lens is converging.
– Use the magnification information (same size implies |m|=1) and image type (inverted implies m=-1) to relate object and image distances (v=u).
– Apply the lens formula (1/f = 1/v – 1/u) with appropriate sign conventions or use the special case rule for an image of the same size formed by a converging lens (object at 2f).
– Determine the sign of the focal length to confirm the nature of the lens (positive f for converging).
– Object at infinity: real, inverted, point image at F.
– Object beyond 2F: real, inverted, diminished image between F and 2F.
– Object at 2F: real, inverted, same size image at 2F.
– Object between F and 2F: real, inverted, magnified image beyond 2F.
– Object at F: real, inverted, image at infinity.
– Object between the lens and F: virtual, erect, magnified image on the same side as the object.
The condition “real, inverted, and of same size” is unique to the object being placed at 2F for a converging lens.