Two similar thin equi-convex lenses, of focal length f each, are kept co-axially in contact with each other such that the focal length of the combination is F₁. When the space between the two lenses is filled with glycerine (which has the same refractive index (μ=1.5) as that of glass) then the equ…

Question

Two similar thin equi-convex lenses, of focal length f each, are kept co-axially in contact with each other such that the focal length of the combination is F₁. When the space between the two lenses is filled with glycerine (which has the same refractive index (μ=1.5) as that of glass) then the equivalent focal length is F₂. The ratio F₁:F₂ will be

Accepted Answer

(1) 1 : 2. Given: two identical thin equi-convex lenses of focal length f in contact, first with air between them and then with glycerine of the same index as the glass. Idea: lenses in contact add their powers, and the glycerine between the two convex faces is itself a lens — a bi-concave one. With air between: 1/(F₁)=1/f+1/f=2/f, so F₁=f/2. For one equi-convex glass lens, 1/f=(1.5-1)2/R, which gives R=f. The glycerine layer has radii -R and +R and the same μ=1.5: 1/(f_g)=(1.5-1)(-2/R)=-1/R, so f_g=-f. Three lenses in contact: 1/(F₂)=1/f-1/f+1/f=1/f, so F₂=f. F₁:F₂=f/2:f=1:2

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