Question:medium

The focal length of a convex lens is $20\text{ cm}$. An object is placed at $40\text{ cm}$ from the lens. The image formed will be:

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Whenever an object is placed at $2F$ ($u = 2f = 40\text{ cm}$) for a convex lens, the image is formed on the other side at $2F$. Like all real images formed by a single lens, it is Real and Inverted.
Updated On: May 30, 2026
  • Virtual and erect
  • Real and inverted
  • Virtual and inverted
  • Real and erect
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The Correct Option is B

Solution and Explanation

Step 1 : Understanding the Question:
The topic of this question is Ray Optics, specifically focusing on Image Formation by Lenses. A convex lens is a converging lens that can form different types of images (real, virtual, magnified, or diminished) depending on the position of the object relative to the focal point ($F$) and the center of curvature ($2F$). The question gives us the focal length and the object distance and asks for the nature of the resulting image.
Step 2 : Key Formulas and approach:
The approach involves identifying the specific "Case" of image formation based on the object distance ($u$) and focal length ($f$):
1. Given: $f = 20\text{ cm}$, $u = 40\text{ cm}$.
2. Relationship: We observe that $u = 2f$ (since $40 = 2 \times 20$).
3. Lens Formula (for confirmation): $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Step 3 : Detailed Explanation:

In optics, $2f$ represents the center of curvature of the lens surface.

Case Analysis: When an object is placed exactly at $2F$ of a convex lens, the rays of light converge on the other side of the lens at exactly the $2F$ position.

Using the lens formula: $\frac{1}{v} - \frac{1}{-40} = \frac{1}{20}$. This simplifies to $\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40}$. Thus, $v = +40\text{ cm}$.

Nature of Image: Since the image distance ($v$) is positive, it means the image is formed on the side opposite to the object. The light rays physically meet at this point, making it a "Real" image.

Real images formed by a single lens are always "Inverted" (upside down).

Size: The magnification $m = v/u = 40/(-40) = -1$. The negative sign confirms it is inverted, and the magnitude of 1 means it is the same size as the object.

Therefore, an object at $2F$ produces a real and inverted image at $2F$.

Step 4 : Final Answer:
Because the object is at $2F$, the image is real and inverted. The correct option is (B).
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