Lenses — Maharashtra (SSC) Class 10 Science Solutions (Free)
Free step-by-step Maharashtra (SSC) Class 10 Science solutions for "Lenses" — important questions with detailed answers, download PDF for board exam preparation.
TL;DR: Free step-by-step Maharashtra (SSC) Class 10 Science solutions for "Lenses" — important questions with detailed answers, download PDF for board exam p…
By Syllab.in · Updated
Q1: Define lens. Classify lenses and state characteristics of each type.
Lens: Transparent optical device made of glass or plastic with curved surfaces that refracts light. Classification: (1) Convex lens (converging): Thicker at center, diverges light outward at edges but converges rays toward focus. Produces real or virtual images. (2) Concave lens (diverging): Thinner at center, thicker at edges, diverges light away from center. Always produces virtual, erect, diminished images. Both have principal focus, focal length, and optical center. Converging lens: +ve foca…
Q2: Define focal length and focal point. State lens maker's formula.
Focal length: Distance from optical center to principal focus where parallel rays converge (convex) or appear to diverge (concave). Unit: meter (m). Focal point: Point where rays converge or appear to diverge. Lens maker's formula: 1/f = (n-1)[1/R1 - 1/R2], where f = focal length, n = refractive index of lens material, R1 and R2 = radii of curvature of two surfaces. Formula relates focal length to refractive index and curvatures of lens surfaces. For symmetric lens, R2 = -R1.
Q3: State and derive lens formula. What does it represent?
Lens formula: 1/f = 1/v + 1/u, where f = focal length, u = object distance (negative from lens), v = image distance (positive/negative from lens). Sign convention: Real objects have negative u, real images have positive v, virtual images have negative v. Formula relates positions and focal length. Derivation uses similar triangles from ray diagrams and geometry of lens. Formula applies to both convex and concave lenses with proper sign convention. Used to find image position given object positio…
Q4: An object 3 cm tall is placed 20 cm from a convex lens of focal length 15 cm. Find image distance and magnification.
Step 1: Given u = -20 cm (object distance), f = 15 cm, h_o = 3 cm
Step 2: Use lens formula: 1/f = 1/v + 1/u
Step 3: 1/15 = 1/v + 1/(-20)
Step 4: 1/15 = 1/v - 1/20
Step 5: 1/v = 1/15 + 1/20 = 4/60 + 3/60 = 7/60
Step 6: v = 60/7 = 8.57 cm
Step 7: Magnification m = -v/u = -(8.57/-20) = 0.43. Image is real, inverted, diminished, at 8.57 cm from lens.
Q5: Define magnification. What are conditions for different types of images formed by convex lens?
Magnification: Ratio of image height to object height. Formula: m = h_i/h_o = -v/u. Positive magnification: Virtual, erect images. Negative magnification: Real, inverted images. Magnification > 1: Image enlarged. Magnification < 1: Image diminished. Conditions for convex lens: (1) Object beyond 2F: Real, inverted, diminished image. (2) Object at 2F: Real, inverted, same size. (3) Object between F and 2F: Real, inverted, enlarged. (4) Object at F: Image at infinity. (5) Object between F and…
Showing 5 of 7 questions — full solutions on the page.