Ray Optics - Lenses and Mirrors Solved Examples (Class 12 Physics)
Ray optics numericals cover lens formula, magnification, mirror equation, and optical instruments. These problems explain image formation and practical app
TL;DR: Ray optics numericals cover lens formula, magnification, mirror equation, and optical instruments. These problems explain image formation and practica…
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Ray optics numericals cover lens formula, magnification, mirror equation, and optical instruments. These problems explain image formation and practical app
Ray Optics - Lenses and Mirrors — Solved Numerical Examples (Step by Step)
Example 1: A convex lens has focal length 15 cm. An object 2 cm tall is placed at 30 cm from the lens. Find image position, magnification, and image height.
Solution: Focal length f = 15 cm, Object distance u = 30 cm, Object height h = 2 cm. Using lens formula 1/f = 1/u + 1/v: 1/15 = 1/30 + 1/v. 1/v = 1/15 - 1/30 = 2/30 - 1/30 = 1/30. So v = 30 cm. Magnification m = -v/u = -30/30 = -1. Image height = m × h = -1 × 2 = -2 cm (real, inverted, same size).
Example 2: A concave mirror has radius of curvature 20 cm. An object 5 cm tall is placed at 15 cm from the mirror. Find image position, magnification, and nature.
Solution: Radius of curvature R = 20 cm, Focal length f = R/2 = 10 cm. Object distance u = 15 cm. Using mirror formula 1/f = 1/u + 1/v: 1/10 = 1/15 + 1/v. 1/v = 1/10 - 1/15 = 3/30 - 2/30 = 1/30. So v = 30 cm. Magnification m = -v/u = -30/15 = -2. Image height = -2 × 5 = -10 cm. Image is real, inverted, magnified.
Example 3: A convex lens of focal length 10 cm produces a virtual image at 5 cm from the lens. Find object distance.
Solution: Focal length f = 10 cm, Image distance v = -5 cm (virtual). Using 1/f = 1/u + 1/v: 1/10 = 1/u + 1/(-5). 1/u = 1/10 + 1/5 = 1/10 + 2/10 = 3/10. So u = 10/3 = 3.33 cm.
Example 4: Two lenses with focal lengths 20 cm and 30 cm are placed in contact. Calculate equivalent focal length.
Solution: f₁ = 20 cm, f₂ = 30 cm. Power of lens 1: P₁ = 1/f₁ = 1/0.2 = 5 D. Power of lens 2: P₂ = 1/f₂ = 1/0.3 = 3.33 D. Equivalent power Peq = P₁ + P₂ = 5 + 3.33 = 8.33 D. Equivalent focal length feq = 1/8.33 = 0.12 m = 12 cm.
Example 5: A person with near point 50 cm wants to read at 25 cm. What is the focal length of the lens required?
Solution: For reading, image must be at near point. Object at 25 cm should form virtual image at 50 cm. Using 1/f = 1/u + 1/v: 1/f = 1/25 + 1/(-50) = 1/25 - 1/50 = 2/50 - 1/50 = 1/50. So f = 50 cm.
Tips
- Lens formula: 1/f = 1/u + 1/v. Magnification m = -v/u = h'/h.
- Real images are inverted (m negative); virtual images are upright (m positive).
- For mirrors: concave mirrors converge light (f positive in formula used); convex mirrors diverge.
Frequently Asked Questions
How do convex and concave lenses differ?
Convex lenses converge parallel light rays to a focal point (real focus). Concave lenses diverge parallel rays as if coming from a virtual focus behind the lens.
Why do spectacles use different lens powers for different people?
Different people have different focal lengths of the eye. Spectacle lenses correct vision by adjusting the focal length of the eye-lens combination to focus images correctly on the retina.
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