Question

A double concave lens with refractive index (n)=1.5 is kept in the air. It's two spherical surfaces have radii R1 =20cm and R2=60cm.Find the focal length of the lens. write the characteristics of the lens.

Answer 1

we know,

lens maker formula,

1/f = (u -1){ 1/R1 - 1/R2 }

where f is focal length of lens
u is the refractive index of lens with respect to medium .
R1 and R2 are the radius of curvature of both curve surface of lens .

given
n = 1.5
so, u = n reative to vacuum or air
u = n/1 = 1.5/1 = 1.5

R1 = -20 cm
R2 = 60 cm

so ,

1/f =(1.5-1) { 1/-20 - 1/60 }

1/f = (0.5)(-4/60)

f = -30 cm

character of concave lens :-
★ focal length of concave lens always negative .

★ image form virtual, erect ,and diminished

★ concave lens always diverging lens .e.g when light rays comes from infinite in lens , then light diverge by lens .

Answer 2

see diagram.

The sign convention for quantities is that distances in the direction of light rays are positive. Other way round are negative.

For refraction on the first concave surface we can derive that:
μ₂/v₁ - μ₁/u = (μ₂-μ₁)/R₁

μ₁/v - μ₂/v₁ = - (μ₂ - μ₁)/R₂

Summing up we get    1/v - 1/u = (μ₂₁ - 1) [1/R₁  - 1/R₂]
       The quantity on LHS is 1/f.
So  Lens maker's equation:   1/f = (μ₂₁ -1) [1/R₁ - 1/R₂]

      R₁ = -20 cm ,   R₂ = +60 cm
      1/f = (1.5 - 1) * [-1/20 - 1/60] = - 1/30
       f = -30 cm

Characteristics:
    The double concave lens diverges rays coming from the object. So the image is always virtual, smaller than the object. The virtual erect diminished image is formed on the same side as the object.
 
    The focal length is negative. The distant objects (far off ones) form a virtual erect point image at the focus, on the same side.

    This lens's diverging property is used to correct nearsightedness (defect) of the eye.