Home › ICSE Class 10 › physics › Refraction through a Lens › An equiconvex thin lens has both surfaces of cur…
An equiconvex thin lens has both surfaces of curvature 20 cm and refractive index 1.5. Its focal length is:
A10 cm
B15 cm
C30 cm
D20 cm
Answer & Solution
Correct answer: D. 20 cm
1. Lens-maker's formula: 1/f = (n - 1) * (1/R1 - 1/R2).
2. For equiconvex lens, R1 = +20 cm and R2 = -20 cm.
3. So 1/R1 - 1/R2 = 1/20 - (-1/20) = 2/20 = 1/10.
4. 1/f = (1.5 - 1) * 1/10 = 0.5 / 10 = 1/20.
5. f = 20 cm.
_Source: Selina Concise Physics ICSE Class 10, Ch 5 'Refraction through a Lens' (aplustopper.com extract)_
Related questions
Linear magnification produced by a converging lens for an object at 2F is:Cross-section of a convex lens looks like:Two lenses of powers +3 D and -5 D are placed in contact. The combination:A virtual image:Image formed by a convex lens with object at infinity is:If both surfaces of a thin lens have equal radii of curvature R, the lens is called:An object is placed beyond 2F from a convex lens. Image position is:Which is the principal use of a concave lens?