Q.45 An object is placed at the principal focus of a concave lens of focal length 10
cm. The image will be formed at _____cm, between the optical center and the
focus of the lens on the same side of the object.
Lens Formula Application
Concave lenses use the standard lens formula \(\frac{1}{v} – \frac{1}{u} = \frac{1}{f}\) with Cartesian sign convention, where distances to the left of the optical center (incident light direction) are negative.
For focal length \(f = -10\) cm and object at principal focus \(u = -10\) cm, substitute values: \(\frac{1}{v} = \frac{1}{-10} + \frac{1}{-10} = -\frac{1}{5}\), so \(v = -5\) cm.
The negative \(v\) confirms a virtual image on the same side as the object, specifically 5 cm left of the optical center—midway between center (0 cm) and focus (-10 cm).
Ray Diagram Explanation
Two key rays trace the image: a ray parallel to principal axis diverges after lens, appearing from focus; a ray through optical center passes undeviated.
Their backward extensions intersect 5 cm left of center, forming erect, diminished virtual image between optical center and focus.
Option Analysis
The blank requires a numeric distance matching “between the optical center and the focus” (0 to 10 cm from center). 5 cm fits exactly (halfway).
Other values like 10 cm (at focus) or 2.5 cm mismatch calculation; infinity or 20 cm imply real images impossible for concave lenses.
Key Characteristics
- Virtual, erect, diminished image always for concave lenses.
- Position between optical center (0 cm) and focus (10 cm).
- Independent of exact object size; scales with focal length.
Practical Exam Tips
- Apply Cartesian convention: f negative, u negative for left-side object.
- Verify with magnification \(m = \frac{v}{u} = 0.5\) (half size).
- Common MCQ trap: confusing with convex lens behavior.


