Question

when two lenses of powers +3D and -2D are kept in cinduct. what is power of the resultant lens?​

Answer 1

Answer: P = 1 D

Explanation:

P (resultant) = Sum of all powers of the associated lens with sign convention

=> P = P1 + P2

=> P = + 3D + (- 2D)

=> P = 1 D.

More:

• 1 dioptre is defined as the reciprocal of focal length equal to 1 m of a lens.
• 'x' D = 1/x m-1 (x in metres) = 100/x cm-1 (x in centimetres); R = 2f.
• P ∝ 1/f. So, graph __ (see attachment).
• Resultant power of a pair of lens kept at a distance d = $\dfrac{f_1 + f_2 - d}{f_1 f_2}$.
• Power of lens required for a myopic eye = $- F$ if F = in metres.
• Power of lens required for a hypermetropic eye = $4 - \dfrac{1}{N.P.}$ if N.P. = near point (in metres).
• Myopic eye requires convex lens and that of hypermetropic one is concave lens.
• Postive power implies that the lens is converging and negative implies that the lens is diverging.