The refractive index of the material of a double equiconvex lens is 1.5.if the radius of a curvature of the lens is r. Then its focal length is?

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Gowtham · Answer

focal length is equal to its radius of curvature from the lens maker formula 1/f=((n2-n1)/n1)(1/r1-1/r2) here n1= refractive index of air n2=refractive index of lens r1 is positive and r2 is negative (from sign convention) so 1/f=1/r f=r

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The refractive index of the material of a double equiconvex lens is 1.5.if the radius of a curvature of the lens is r. Then its focal length is?

The refractive index of the material of a double equiconvex lens is 1.5.if the radius of a curvature of the lens is r. Then its focal length is?

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The refractive index of the material of a double equiconvex lens is 1.5.if the radius of a curvature of the lens is r. Then its focal length is?

The refractive index of the material of a double equiconvex lens is 1.5.if the radius of a curvature of the lens is r. Then its focal length is?

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