Question

A concave lens forms an image that is 1/3rd the size of the object, as shown in the diagram. If the object is placed 24 cm in front of the lens, determine:
A. the type of lens.
B. the image distance
C. the focal length of the lens

Hint:

The image is erect, so \(m = +1/3\). Use \(v = mu\) and then the lens formula.

Answer 1

Step 1: Part A: Identify the type of lens:
The figure shows a lens that is thinner at the centre and thicker at the edges. The ray parallel to the axis leaves the lens as if it came from the focus \(F_1\) on the object side. The ray through the optical centre goes straight. These rays spread out. A lens that spreads parallel rays is a concave (diverging) lens. It always forms a virtual, erect and diminished image. So the lens is concave.

Step 2: Part B: Set up the magnification:
Magnification \(m = \dfrac{h_i}{h_o} = \dfrac{v}{u}\). The image is 1/3rd the size of the object. A concave lens gives an erect image, so \(m\) is positive:
\[ m = +\frac{1}{3} \]
The object is 24 cm in front of the lens. By the sign convention, \(u = -24\) cm.

Step 3: Find the image distance:
\[ v = m \times u = \frac{1}{3} \times (-24) = -8 \text{ cm} \]
The negative sign shows that the image is on the same side of the lens as the object. It is virtual and lies 8 cm from the lens.

Step 4: Part C: Use the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Put \(v = -8\) cm and \(u = -24\) cm:
\[ \frac{1}{f} = \frac{1}{-8} - \frac{1}{-24} = -\frac{1}{8} + \frac{1}{24} = \frac{-3 + 1}{24} = -\frac{2}{24} = -\frac{1}{12} \]

Step 5: Read the focal length:
\[ f = -12 \text{ cm} \]
The negative sign again confirms a concave lens. The image at 8 cm lies between the focus (12 cm) and the lens, which matches the figure where \(B'\) is between \(F_1\) and \(O\).

Final Answer:
The lens is concave, the image distance is \(-8\) cm and the focal length is \(-12\) cm. \[ \boxed{\text{Concave lens}, \; v = -8 \text{ cm}, \; f = -12 \text{ cm}} \]