Question

The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20 cm. The focal length of the lens used is __________ cm.

Answer 1

Correct Answer: 15

Let the object distance be \(u\), and the image distance be \(v\). Since the magnification \(m = -\frac{v}{u} = 3\), we have:

\[ v = 3u \]

From the lens formula:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Substituting \(v = 3u\):

\[ \frac{1}{f} = \frac{1}{3u} - \frac{1}{u} \]

Simplifying:

\[ \frac{1}{f} = \frac{1 - 3}{3u} = -\frac{2}{3u} \]

Now, since \(u = 10 \, \text{cm}\), we get:

\[ f = 15 \, \text{cm} \]

Thus, the correct answer is 15.

Answer 2

The problem asks for the focal length of a convex lens that produces a virtual image magnified 3 times, with a separation of 20 cm between the object and the image.

Concept Used:

We will use the lens formula and the magnification formula, applying the standard Cartesian sign convention.

Lens Formula:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Magnification Formula:

\[ m = \frac{v}{u} \]

Sign Convention:

1. All distances are measured from the optical center.
2. Distances measured in the direction of incident light are positive, and those measured against it are negative.
3. The object distance (\(u\)) for a real object is always negative.
4. The focal length (\(f\)) for a convex lens is positive.
5. For a virtual image formed by a convex lens, the image is on the same side as the object, so the image distance (\(v\)) is negative. The magnification (\(m\)) is positive for a virtual, erect image.

Step-by-Step Solution:

Step 1: Analyze the given information based on the sign convention.

The image is virtual and magnified 3 times. Since virtual images are erect, the magnification is positive.

\[ m = +3 \]

Using the magnification formula:

\[ m = \frac{v}{u} \implies +3 = \frac{v}{u} \implies v = 3u \]

Step 2: Use the given distance between the object and the image.

A convex lens forms a virtual image when the object is placed between the optical center and the focal point. This virtual image is formed on the same side as the object. Let the object be placed at a distance \(|u|\) from the lens. According to the sign convention, \(u = -|u|\). The image is formed at a distance \(|v|\) on the same side, so \(v = -|v|\).

The distance between the object and the image is given as 20 cm. Since they are on the same side and the image is magnified, the image is farther from the lens than the object.

\[ |v| - |u| = 20 \text{ cm} \]

Using \(v=3u\), we get \(|3u| - |u| = 20\), which simplifies to \(3|u| - |u| = 20\).

\[ 2|u| = 20 \implies |u| = 10 \text{ cm} \]

Step 3: Determine the object distance (\(u\)) and image distance (\(v\)).

From Step 2, the magnitude of the object distance is 10 cm. Applying the sign convention:

\[ u = -10 \text{ cm} \]

Now, we can find the image distance using the relation from Step 1:

\[ v = 3u = 3 \times (-10 \text{ cm}) = -30 \text{ cm} \]

The negative sign for \(v\) confirms that the image is virtual and on the same side as the object.

Step 4: Calculate the focal length using the lens formula.

Substitute the values of \(u\) and \(v\) into the lens formula:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] \[ \frac{1}{f} = \frac{1}{-30} - \frac{1}{-10} \] \[ \frac{1}{f} = -\frac{1}{30} + \frac{1}{10} \]

Final Computation & Result:

To find \(f\), we compute the sum on the right side:

\[ \frac{1}{f} = \frac{-1 + 3}{30} = \frac{2}{30} = \frac{1}{15} \]

Therefore, the focal length of the lens is:

\[ f = 15 \text{ cm} \]

The positive value of \(f\) is consistent with a convex lens. The focal length of the lens used is 15 cm.