A convex lens has a focal length of 15 cm. At what distance should an object be placed in front of the lens to get a real image of the same size of the object?
30 cm
Understanding Image Formation by a Convex Lens
The question asks about the specific position where an object should be placed in front of a convex lens to form a real image that is exactly the same size as the object itself. We are given the focal length of the convex lens.
Given Information:
- Type of lens: Convex lens
- Focal length of the lens, \(f = 15 \text{ cm}\)
- Desired image characteristic: Real image of the same size as the object
We need to find the object distance, denoted by \(u\), from the lens.
Condition for Same Size Real Image
For a convex lens, a real image is formed on the opposite side of the lens from the object. When the image formed is real and of the same size as the object, this occurs at a specific object position.
The magnification, \(m\), produced by a lens is given by the ratio of the image height (\(h_i\)) to the object height (\(h_o\)), and also by the ratio of the image distance (\(v\)) to the object distance (\(u\)). The formula is:
\(m = \frac{h_i}{h_o} = \frac{v}{u}\)
For a real image formed by a convex lens, the image is inverted. If the image is also the same size as the object (\(h_i = h_o\)), then the magnitude of magnification is 1. Since the image is real and inverted, the magnification is negative:
\(m = -1\)
Using the magnification formula, we have:
\(-1 = \frac{v}{u}\)
This implies that the image distance \(v\) and the object distance \(u\) have the same magnitude but opposite signs when following standard sign conventions where \(u\) is negative for an object on the left and \(v\) is positive for a real image on the right. So, \(|v| = |u|\). For convenience in the lens formula using magnitudes, we can use the relationship directly derived from \(m=-1\) and the lens formula.
Applying the Lens Formula
The lens formula relates the focal length (\(f\)), object distance (\(u\)), and image distance (\(v\)):
\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
According to the sign convention, for an object placed in front of the lens (typically on the left), \(u\) is negative. For a real image formed by a convex lens (on the right side), \(v\) is positive. The focal length of a convex lens \(f\) is positive.
From the magnification condition for a same-size real image, \(m=-1\), we found that \(v = -u\). Let's substitute \(v = -u\) into the lens formula, keeping in mind the signs. If \(u\) represents the position coordinate (negative), then \(v\) is the position coordinate (positive), and \(v = -u\) means their magnitudes are equal, i.e., \(|v| = |u|\).
Let's work with the magnitudes. For a real image formed by a convex lens, the lens formula can be written using magnitudes as:
\(\frac{1}{f} = \frac{1}{|v|} + \frac{1}{|u|}\)
Since we need a same-size image (\(|m|=1\)), we have \(\frac{|v|}{|u|} = 1\), which means \(|v| = |u|\). Let's substitute \(|v| = |u|\) into the magnitude lens formula:
\(\frac{1}{f} = \frac{1}{|u|} + \frac{1}{|u|}\)
\(\frac{1}{f} = \frac{2}{|u|}\)
Rearranging this equation to find \(|u|\):
\(|u| = 2f\)
This shows that to obtain a real image of the same size as the object using a convex lens, the object must be placed at a distance equal to twice its focal length from the lens. The image will also be formed at a distance of \(2f\) on the other side of the lens.
Calculating the Object Distance
The focal length of the convex lens is given as \(f = 15 \text{ cm}\). Using the derived condition, the object distance should be:
\(|u| = 2 \times f\)
\(|u| = 2 \times 15 \text{ cm}\)
\(|u| = 30 \text{ cm}\)
So, the object should be placed at a distance of 30 cm in front of the convex lens.
Summary of Image Formation by Convex Lens at Different Positions
Understanding how the image characteristics change with object position is crucial. Here's a quick summary:
| Object Position | Image Position | Nature of Image | Size of Image |
|---|---|---|---|
| At infinity | At focus F\(_2\) | Real, inverted | Highly diminished (point size) |
| Beyond 2F\(_1\) | Between F\(_2\) and 2F\(_2\) | Real, inverted | Diminished |
| At 2F\(_1\) | At 2F\(_2\) | Real, inverted | Same size |
| Between F\(_1\) and 2F\(_1\) | Beyond 2F\(_2\) | Real, inverted | Enlarged |
| At Focus F\(_1\) | At infinity | Real, inverted | Highly enlarged |
| Between optical center and F\(_1\) | On the same side as object | Virtual, erect | Enlarged |
From the table, it is clear that when the object is placed at 2F\(_1\), the real, inverted image is formed at 2F\(_2\) and is of the same size as the object. Since F\(_1\) and F\(_2\) are at a distance \(f\) from the lens, 2F\(_1\) is at a distance \(2f\) from the lens.
Therefore, for a convex lens with a focal length of 15 cm, the object must be placed at a distance of \(2 \times 15 \text{ cm} = 30 \text{ cm}\) to get a real image of the same size.
Revision Table: Convex Lens Key Concepts
| Concept | Description | Formula/Condition |
|---|---|---|
| Focal Length (f) | Distance from lens center to focal point. Positive for convex lens. | Given (15 cm) |
| Object Distance (u) | Distance from lens center to object. Negative in sign convention. | To be found |
| Image Distance (v) | Distance from lens center to image. Positive for real image. | Related to u and f |
| Lens Formula | Relates f, u, and v. | \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) |
| Magnification (m) | Ratio of image height to object height; also v/u. | \(m = \frac{v}{u}\) |
| Real Image | Formed by actual intersection of light rays; can be projected. Inverted by convex lens. | v > 0 (typically) |
| Same Size Image | Image height equals object height. | \(|m| = 1\) |
| Same Size Real Image Condition | Occurs when object is placed at \(2f\). | \(|u| = 2f\), \(|v| = 2f\), \(m = -1\) |
Additional Information: Optics and Ray Diagrams
Understanding image formation is often helped by drawing ray diagrams. For a convex lens, principal rays are used:
- A ray parallel to the principal axis passes through the focal point F\(_2\) after refraction.
- A ray passing through the focal point F\(_1\) becomes parallel to the principal axis after refraction.
- A ray passing through the optical center goes undeviated.
When the object is placed at a distance \(2f\) from the convex lens, drawing these rays shows that they intersect at a distance \(2f\) on the other side, forming an inverted image of the same height. This position, 2F\(_1\), is also called the center of curvature in the case of a lens formed by two surfaces of radius R = 2f (thin lens approximation).
This scenario (\(|u| = 2f, |v| = 2f, m = -1\)) is a critical point in the characteristics of image formation by a convex lens, marking the transition from a diminished image (object beyond \(2f\)) to an enlarged image (object between \(f\) and \(2f\)).
Important Questions from Mirrors and Images
- Which of the following statements correctly describes the nature and position of the image formed by a convex mirror when a real object is placed at any position in front of it?
- A beam of parallel light, originating from a distant source, is first incident on a convex lens with focal length $f_2$.
Subsequently, the light passes through the lens and then reflects from a concave mirror having a focal length $f_1$.
The concave mirror is placed at a distance $d$ from the convex lens.
For the light rays to retrace their original path and ultimately emerge from the lens as a parallel beam heading back towards the distant source, the separation distance $d$ between the lens and the mirror must be: The total number of images formed by two mirrors inclined at 72° to each other when the object is placed unsymmetrically will be ___?
A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from the pole of the mirror. The size of the image is ;
- A concave mirror of focal length $f$ produces an image $n$ times the size of the object. If the image is virtual, then the distance of the object from the mirror is:
