An equiconcave lens is made of glass with a refractive index of 1.5 . If the radius of curvature of each of its surfaces is 20 cm , what is the power of the lens?

Step-by-step solution

For an equiconcave lens, the radii of curvature are R₁ = -20 cm and R₂ = +20 cm . Using the Lens Maker's formula: 1 f = ( - 1) ( 1 R₁ - 1 R₂ ) Substitute the given values: 1 f = (1.5 - 1) ( 1 -20 - 1 20 ) 1 f = 0.5 (- 2 20 ) = - 1 20 cm ⁻¹ The focal length is f = -20 cm = -0.2 m . The power of the lens is given by P = 1 f( in m ) . P = 1 -0.2 = -5 D . Answer: -5 D