Question

Two identical equiconvex lens L₁ and L₂ , made of a material of refractive index μ₁ = 1.5 and having a focal length, f, each, are placed in contact, co–axially, as shown in Fig(a). The space between the two lenses, is sealed at the two ends, after being filled with a transparent liquid of referactive index, μ₂ = 1.8, as shown in Fig(b). The change, in the equivalent focal length, of the arrangment, is

(1) + f

(2) – 1.5 f

(3) + 2.0 f

(4) – 2.0 f

Answer 1

(3) + 2.0 f

The focal length, of each lens, in terms of the equal radii of curvature, c, of its two surfaces, is

The equivalent focal length, F₁ , of the combination shown in Fig(a), is

The arrangment, shown in Fig(b), is a combination of three thin lenses. The additional lens is an equiconcave liquid lens. Let f3 be the focal length of liquid lens. Then

The equivalent focal length, F₁ , of the arrangment shown in Fig(b), is

The change in focal length = F₂ - F₁ = 2.5 f – 0.5 f 

= + 2.0 f