Question

An object placed in front of a convex lens, forms an image of same size on a screen. Moving the object 12 cm closer to the lens results in the formation of a real image which is three times the size of the object. Calculate the focal length of the lens.

Hint:

The condition "image of same size" for a convex lens is a strong hint that the object is at 2f. This is often the starting point for solving such problems.

Answer 1

Step 1: Analyze the initial condition.
A convex lens forms a real image of the same size as the object only when the object is placed at a distance of twice the focal length (2f) from the optical center. Initial object distance, \(u = -2f\). The image is also formed at 2f on the other side.
Step 2: Analyze the second condition.
The object is moved 12 cm closer to the lens. New object distance, \(u' = u + 12 = -2f + 12\). (The negative sign is included in the value).
A real image is formed, which is three times the size. For a real image from a convex lens, the magnification (\(m\)) is negative (inverted). Magnification, \(m = -3\).
Step 3: Use the magnification formula.
\(m = \frac{v'}{u'}\) \[ -3 = \frac{v'}{-2f + 12} \] \[ v' = -3(-2f + 12) = 6f - 36 \] This is the new image distance. Since \(v'\) is positive, the image is real, which matches the problem description.
Step 4: Use the lens formula.
\[ \frac{1}{f} = \frac{1}{v'} - \frac{1}{u'} \] Substitute the expressions for \(u'\) and \(v'\): \[ \frac{1}{f} = \frac{1}{6f - 36} - \frac{1}{-2f + 12} \] \[ \frac{1}{f} = \frac{1}{6f - 36} + \frac{1}{2f - 12} \] Find a common denominator: \[ \frac{1}{f} = \frac{1}{3(2f - 12)} + \frac{1}{2f - 12} \] \[ \frac{1}{f} = \frac{1 + 3}{3(2f - 12)} \] \[ \frac{1}{f} = \frac{4}{3(2f - 12)} \] Cross-multiply: \[ 3(2f - 12) = 4f \] \[ 6f - 36 = 4f \] \[ 2f = 36 \] \[ f = 18 \text{ cm} \] The focal length of the lens is 18 cm.