An object is placed on the principal axis of a convex lens of focal length 10 cm. If the distance of the object from the lens is 30 cm, what is the distance of the image formed?
Calculating Image Distance with a Convex Lens
This problem requires us to find the distance of the image formed by a convex lens when we know the object distance and the focal length. We can use the lens formula, which relates these three quantities.
Understanding the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]Where:
- \(f\) is the focal length of the lens.
- \(v\) is the image distance (distance of the image from the lens).
- \(u\) is the object distance (distance of the object from the lens).
Applying Sign Convention for Convex Lens
To use the lens formula correctly, we must apply the standard sign convention:
- Distances measured in the direction of incident light are taken as positive.
- Distances measured in the direction opposite to the incident light are taken as negative.
- The focal length of a convex lens is positive.
- Object distance \(u\) is usually taken as negative because the object is placed on the left side (opposite to the direction of incident light originating from the object).
From the question, we are given:
- Focal length of the convex lens, \(f = +10\) cm (positive for a convex lens).
- Object distance from the lens, \(u = -30\) cm (assuming the object is placed on the left).
Step-by-Step Calculation of Image Distance
Now, let's substitute the given values into the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] \[ \frac{1}{10} = \frac{1}{v} - \frac{1}{-30} \] \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{30} \]To find \(v\), we need to isolate \(\frac{1}{v}\):
\[ \frac{1}{v} = \frac{1}{10} - \frac{1}{30} \]Find a common denominator, which is 30:
\[ \frac{1}{v} = \frac{3}{30} - \frac{1}{30} \] \[ \frac{1}{v} = \frac{3 - 1}{30} \] \[ \frac{1}{v} = \frac{2}{30} \] \[ \frac{1}{v} = \frac{1}{15} \]Now, take the reciprocal of both sides to find \(v\):
\[ v = 15 \text{ cm} \]Interpreting the Result
The calculated image distance \(v\) is +15 cm. The positive sign indicates that the image is formed on the right side of the lens (the side where light exits the lens), which is characteristic of a real image formed by a convex lens when the object is placed beyond the focal length.
Therefore, the distance of the image formed is 15 cm.
| Quantity | Symbol | Value | Sign Convention |
|---|---|---|---|
| Focal Length (Convex Lens) | \(f\) | 10 cm | Positive (+) |
| Object Distance | \(u\) | 30 cm | Negative (-) |
| Image Distance | \(v\) | ? | To be calculated |
Revision Table: Convex Lens Concepts
| Concept | Description |
|---|---|
| Convex Lens | Also known as a converging lens, it is thicker at the center than at the edges and converges parallel rays of light to a point (the focal point). |
| Focal Length (\(f\)) | The distance from the optical center of the lens to the principal focus. It is positive for a convex lens. |
| Principal Axis | The straight line passing through the optical center and perpendicular to the lens surface. |
| Object Distance (\(u\)) | Distance of the object from the optical center of the lens. Taken as negative when the object is placed on the left side (standard convention). |
| Image Distance (\(v\)) | Distance of the image from the optical center of the lens. Positive \(v\) means a real image on the right side; negative \(v\) means a virtual image on the left side. |
| Lens Formula | \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) (for thin lenses). |
Additional Information: Image Formation by Convex Lens
The nature and position of the image formed by a convex lens depend on the position of the object relative to the lens and its focal point. In this problem, the object is placed at 30 cm, which is beyond 2 times the focal length (\(2f = 2 \times 10 = 20\) cm). When an object is placed beyond \(2f\) of a convex lens, the image formed is:
- Real
- Inverted
- Diminished
- Formed between \(f\) and \(2f\) on the other side of the lens.
Our calculated image distance \(v = 15\) cm is between \(f = 10\) cm and \(2f = 20\) cm, which is consistent with the expected image location for an object placed beyond \(2f\).
Important Questions from Optics
Which one of the following colours may be obtained by combining green and red colours?
Which of the following are the primary colours of light?
Directions: The following items consist of two statements, Statement I and Statement II. You are to examine these two statements carefully and select the answers to these items using the code given below:
Statement I: Diamond is very bright.
Statement II: Diamond has very low refractive index
A non-SI unit called 'nit' is the unit of which of the following photometric quantities used to measure a multitude of light intensity?
Which among the following is used as a reflector in search lights?
