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Question

# Draw a ray diagram showing the formation of the image by a point object on the principal axis of a spherical convex surface separating two media of refractive indices n1 and n2, when a point source is kept in rarer medium of refractive index n1. Derive the relation between object and image distance in terms of refractive index of the medium and radius of curvature of the surface.Hence obtain the expression for lens-maker's formula in the case of thin convex lens.

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Solution

## Consider a convex lens (or concave lens) of absolute refractive index μ2 to be placed in a rarer medium of absolute refractive index μ1.Considering the refraction of a point object on the surface XP1Y, the image is formed at I1 who is at a distance of V1.CI1=P1I1=V1 (as the lens is thin)CC1=P1C1=R1CO=P1O=uIt follows from the refraction due to convex spherical surface XP1Yμ1−u+μ2v1=μ2−μ1R1..........(i)The refracted ray from A suffers a second refraction on the surface XP2Y and emerges along BI. Therefore I is the final real image of O.Here the object distance isu=CI1≈P2I1=VP1P2 is very smallLet CI≈P2I=V(Final image distance)Let R2 be radius of curvature of second surface of the lens.It follows from refraction due to concave spherical surface from denser to rarer medium that−μ2v1+μ1v=μ1−μ2R2=μ2−μ1−R2............(ii)Adding (i)and(ii)−μ1u+μ1v=(μ2−μ1)(1R1−1R2)orμ1(1v−1u)=(μ2−μ2)(1R1−1R2)But 1v−1u=1f and μ2μ1=μThus we get=1f=(μ−1)(1R1−1R2)

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