Light – Reflection & Refraction: Solved Board Numericals & Ray Diagram Q&A
Step-by-step solved numericals on mirror formula, lens formula, power of lenses, and refractive index for CBSE & ICSE Class 10 Physics.
Light Reflection & Refraction – Solved Board Numericals
A concave lens has a focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also, find the magnification produced by the lens.
1. Given values with Cartesian sign convention:
• Focal length of concave lens, f = -15 cm
• Image distance (virtual image on object side), v = -10 cm
• Object distance, u = ?
2. Using the Lens Formula:
1/v - 1/u = 1/f
1/u = 1/v - 1/f
1/u = 1/(-10) - 1/(-15)
1/u = -1/10 + 1/15 = (-3 + 2) / 30 = -1/30
u = -30 cm
3. Magnification calculation:
m = v / u = (-10 cm) / (-30 cm) = +1/3 = +0.33
Final Answer: The object must be placed at a distance of 30 cm in front of the concave lens. The positive sign of magnification indicates a virtual, erect, and diminished image (one-third the size of the object).
A doctor prescribes a corrective lens of power +2.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging? What defect of vision does it correct?
Focal length calculation:
P = 1 / f (in metres)
f = 1 / P = 1 / (+2.5 D) = +0.4 m = +40 cm
Inferences:
• Since focal length and power are positive, it is a Convex (Converging) lens.
• A convex lens corrects Hypermetropia (Farsightedness).
Indexed Topics & Examination Keywords
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