Question

The distance between object and its two times magnified real image as produced by a convex lens is 45 cm. The focal length of the lens used is ______ cm.

Answer 1

Correct Answer: 10

Step 1. Understanding the Given Condition: Since the image is real, inverted, and twice the size of the object, we know:

\( m = \frac{v}{u} = -2 \Rightarrow v = -2u \)

Step 2. Set up Equation Using Total Distance: The distance between the object and the image is 45 cm, so:

\( |v - u| = 45 \, \text{cm} \)

Substitute \( v = -2u \) into the equation:

\( |-2u - u| = 45 \)

\( 3|u| = 45 \Rightarrow u = -15 \, \text{cm} \)

Step 3. Determine Image Distance \( v \): Using \( v = -2u \):

\( v = -2 \times (-15) = 30 \, \text{cm} \)

Step 4. Calculate Focal Length Using Lens Formula: Apply the lens formula:

\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

Substitute \( u = -15 \, \text{cm} \) and \( v = 30 \, \text{cm} \):

\( \frac{1}{f} = \frac{1}{30} - \frac{1}{-15} = \frac{1}{30} + \frac{1}{15} = \frac{1 + 2}{30} = \frac{3}{30} = \frac{1}{10} \)

\( f = +10 \, \text{cm} \)

Answer 2

The problem asks for the focal length of a convex lens that produces a real image, magnified two times, with a distance of 45 cm between the object and the image.

Concept Used:

The solution involves the use of the lens formula and the magnification formula for a thin lens. The standard Cartesian sign convention is used, where the optical center is the origin.

1. Lens Formula: The relationship between the object distance (\(u\)), image distance (\(v\)), and focal length (\(f\)) is given by:

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

2. Magnification Formula: The linear magnification (\(m\)) is given by:

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

For a real image, the magnification is negative (since the image is inverted). For a convex lens, the focal length (\(f\)) is positive, the object distance (\(u\)) is negative, and for a real image, the image distance (\(v\)) is positive.

Step-by-Step Solution:

Step 1: Determine the magnification and relate the image and object distances.

The image is real and magnified two times. A real image formed by a single lens is always inverted. Therefore, the magnification \(m = -2\).

Using the magnification formula:

\[ -2 = \frac{v}{u} \implies v = -2u \]

This relationship is consistent with our sign convention: since \(u\) is negative, \(v\) will be positive.

Step 2: Use the given distance between the object and the image to form a second equation.

The object is placed to the left of the lens (\(u < 0\)), and the real image is formed to the right (\(v > 0\)). The distance between them is the sum of the magnitudes of their positions.

Distance = \(|v| + |u|\). Since \(v > 0\) and \(u < 0\), this is \(v - u\).

\[ v - u = 45 \, \text{cm} \]

Step 3: Solve the two equations to find the object distance (\(u\)) and image distance (\(v\)).

We have a system of two linear equations:

1. \(v = -2u\)
2. \(v - u = 45\)

Substitute the first equation into the second:

\[ (-2u) - u = 45 \] \[ -3u = 45 \] \[ u = -15 \, \text{cm} \]

Now, substitute the value of \(u\) back into the first equation to find \(v\):

\[ v = -2(-15) = 30 \, \text{cm} \]

So, the object is 15 cm in front of the lens, and the image is formed 30 cm behind the lens.

Step 4: Use the lens formula to calculate the focal length (\(f\)).

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

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

Final Computation & Result:

Find a common denominator to add the fractions:

\[ \frac{1}{f} = \frac{1}{30} + \frac{2}{30} \] \[ \frac{1}{f} = \frac{3}{30} = \frac{1}{10} \]

Therefore, the focal length is:

\[ f = 10 \, \text{cm} \]

The focal length of the convex lens is 10 cm.