Question

The focal length of a convex lens is 18 cm and the size of the image is a quarter of the object. The object is situated at a distance of:

• A. \(90 \text{ cm}\)
• B. \(54 \text{ cm}\)
• C. \(22.5 \text{ cm}\)
• D. \(60 \text{ cm}\)

Hint:

When solving lens problems: 1. List knowns (\(f, u\) or \(v, m\)) with correct signs. 2. Use the lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) and magnification \(m = v/u\). If a problem's text and a given "correct" option lead to different physical outcomes (like different magnifications), it's important to note the inconsistency. For this problem, an object at \(54 \text{ cm}\) yields \(m=-1/2\), while \(m=-1/4\) (text) implies \(u=-90 \text{ cm}\).

Answer 1

The Correct Option is B

Concept: This problem involves the lens formula and magnification for a convex lens.
Lens Formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
Magnification Formula: \( m = \frac{v}{u} \) Where \(f\) is focal length, \(u\) is object distance, \(v\) is image distance, and \(m\) is magnification. Sign Convention for convex lens:
\(f\) is positive.
\(u\) is negative (object typically to the left).
\(v\) is positive for a real image.
\(m\) is negative for a real, inverted image. Step 1: Given information and selected option
Focal length, \(f = +18 \text{ cm}\).
The selected correct answer states the object distance is \(54 \text{ cm}\). So, we take \(u = -54 \text{ cm}\). The problem text also states: "the size of the image is a quarter of the object." For a convex lens, a diminished image implies it's real and inverted, so this would mean \(m = -1/4\). We will check this condition later. Step 2: Calculate image distance (\(v\)) using \(u = -54 \text{ cm}\) Using the lens formula: \[ \frac{1}{18} = \frac{1}{v} - \frac{1}{-54} \] \[ \frac{1}{18} = \frac{1}{v} + \frac{1}{54} \] \[ \frac{1}{v} = \frac{1}{18} - \frac{1}{54} \] To subtract, find a common denominator (54): \( \frac{1}{18} = \frac{3}{54} \). \[ \frac{1}{v} = \frac{3}{54} - \frac{1}{54} = \frac{2}{54} = \frac{1}{27} \] So, image distance \(v = +27 \text{ cm}\). The positive sign means the image is real. Step 3: Calculate magnification (\(m\)) for \(u = -54 \text{ cm}\) Using the magnification formula: \[ m = \frac{v}{u} = \frac{27 \text{ cm}}{-54 \text{ cm}} = -\frac{1}{2} \] Step 4: Interpretation of the result and comparison with problem statement The calculated magnification is \(m = -1/2\). This means the image is inverted (due to the negative sign) and its size is half the size of the object. However, the problem statement says, "the size of the image is a quarter of the object" (which implies \(m = -1/4\)). If we were to use \(m = -1/4\), then \(v = mu = (-1/4)u\). Substituting into the lens formula: \( \frac{1}{18} = \frac{1}{(-u/4)} - \frac{1}{u} = -\frac{4}{u} - \frac{1}{u} = -\frac{5}{u} \). This gives \(u = -5 \times 18 = -90 \text{ cm}\). This shows a discrepancy: if the object distance is \(54 \text{ cm}\) (Option 2), the image is half the object's size. If the image is a quarter of the object's size (as per text), the object distance should be \(90 \text{ cm}\) (Option 1). Conclusion based on selected option: Assuming Option (2) \(54 \text{ cm}\) is the intended answer for the object distance, the resulting magnification is \(m = -1/2\).