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Question

# (a) Derive the mathematical relation between refractive indices n1 and n2 of two media and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point since lying on the principle axis in rarer medium of refractive index n1 and a real image formed in the denser medium of refractive index n2. Hence, derive Lens Maker's formula. (b) Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed ?

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Solution

## (a) tan∠NOM=MNOM; tan∠NCM=MNMC; tan∠NIM=MNMI; For △NOC, i is the exterior angle. Assuming the incident ray is very close to the principal axis, all the angles are very small. Hence, for very small angles, tan x=x=sin x ∴i=∠NOM+∠NCM i=MNOM+MNMC ...(i) Similarly,r=∠NCM−∠NIM i.e., r=MNMC−MNMI ...(ii) By Snell's law, n1 sin i=n2 sin r For small angles, n1 i=n2 r On Substituting these values of i and r in equations, we get n1(MNOM+MNMC)=n2(MNMC−MNMI) n1OM+n2MI=n2−n1MC ...(iii) On applying new Cartesian sign conventions, OM=−u,MI=+v,MC=+R Substituting these values in equation (iii),we get n2v−n1u=n2−n1R ...(iv) In deriving Lens maker's formula, we adopt the coordinate geometry sign convention and make the assumptions : (i) The lens is so thin, that the distances measured from the poles of its two surfaces can be taken as equal to the distances from its optical centre. (ii) The aperture of the lens is small. (iii)The object is a point-object placed on the principal axis of the lens. (iv) The incident and the refracted rays make small angles with the principal axis. Refraction at the first surface, n2v′−n1u=n2−n1R1 Refraction at the second surface, n1v−n2v′−t=n1−n2R2 The lens is 'thin', hence t < < v' and can be ignored. Then, we have n1v−n2v′=n1−n2R2 ...(ii) Adding equation (i) and (ii), we get n1v−n2u=n2−n1(1R1−1R2) Putting n2n1=n,the refractive index of the material of the lens with respect to the surrounding medium, we have 1v−1u=(n−1)(1R1−1R2) ...(iii) When the object is at infinity, the image will be formed at the principal focus of the lens, i.e., when u=∞, 1f−1∞=(n−1)(1R1−1R2) 1f=(n−1)(1R1−1R2) This is the Lens Maker's formula. (b) Given,refractive index,n2=1.5,n1=1(air) Radius of curvature,R=20 cm Object distance,u=-100 cm To find,Image distance,v n2v−n1u=n2−n1R 1.5v+1(100)=1.5−120=140 1.5v=140−1100=5−2200=3200 v=2003×1.5=100 cm The image is formed at 100 cm in denser medium.

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