Question

A thin convex lens forms a virtual image when an object is placed on the principal axis at a distance of 12 cm from the lens. When the object is moved away from the lens along the principal axis through a distance of 18 cm, the lens forms a real image. If the magnification of the virtual image is five times the magnification of the real image, then the focal length of the lens is

• A. 27 cm
• B. 18 cm
• C. 15 cm
• D. 12 cm

Hint:

For a convex lens: \[ m=\frac{v}{u} \] and \[ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \] Many lens problems can be solved directly by combining these two relations.

Answer 1

The Correct Option is B

Concept: For a thin convex lens, \[ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \] and magnification is given by \[ m=\left|\frac{v}{u}\right| \] A virtual image formed by a convex lens is erect and lies on the same side as the object, whereas a real image is formed on the opposite side of the lens.

Step 1: Write the object distances in the two cases.
Initially, \[ u_1=-12\text{ cm} \] The object is moved 18 cm away from the lens. Therefore, \[ u_2=-(12+18) \] \[ u_2=-30\text{ cm} \]

Step 2: Find magnification for the virtual image.
Using the lens formula, \[ \frac{1}{f}=\frac{1}{v_1}+\frac{1}{12} \] Solving, \[ v_1=\frac{12f}{12-f} \] Hence, \[ m_1=\frac{v_1}{12} \] \[ m_1=\frac{f}{12-f} \]

Step 3: Find magnification for the real image.
Similarly, \[ \frac{1}{f}=\frac{1}{v_2}+\frac{1}{30} \] \[ v_2=\frac{30f}{30-f} \] Thus, \[ m_2=\frac{v_2}{30} \] \[ m_2=\frac{f}{30-f} \]

Step 4: Use the condition given in the question.
The magnification of the virtual image is five times the magnification of the real image. \[ m_1=5m_2 \] \[ \frac{f}{12-f} = 5\left(\frac{f}{30-f}\right) \] Cancelling \(f\), \[ 30-f=5(12-f) \] \[ 30-f=60-5f \] \[ 4f=30 \] \[ f=18\text{ cm} \]

Step 5: State the answer.
\[ \boxed{f=18\text{ cm}} \]