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Question
With the help of a ray diagram, obtain the relation between its focal length and radius of curvature.
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Solution 1
The distance between the centre of a lens or curved mirror and its focus.
The relationship between the focal length f and the radius of curvature R = 2f.
Consider a ray of light AB, parallel to the principal axis and incident on a spherical mirror at point B. The normal to the surface at point B is CB, and CP = CB = R is the radius of curvature. The ray AB, after reflection from a mirror, will pass through F (concave mirror) or will appear to diverge from F (convex mirror) and obeys the law of reflection, i.e., i = r.
From the geometry of the figure,
∠BCP = θ = i
In DCBF, θ = r
∴ BF = FC (because i = r)
If the aperture of the mirror is small, B lies close to P,
∴ BF = PF
Or, FC = FP = PF
Or, PC = PF + FC = PF + PF
Or, R = 2 PF = 2f
Or, f = R/2
A similar relation holds for a convex mirror also. In deriving this relation, we have assumed that the aperture of the mirror is small.
Solution 2
A ray of light BP' travelling parallel to the principal axis PC is incident on a spherical mirror PP'. It reflects along P'R.
For a concave mirror, it passes through the focus. For a convex mirror, extending the ray backward, it appears to pass through the focus.
P is the pole and Fis the focus of the mirror.
PF = f
C is the centre of curvature.
PC = radius of curvature = R
P'C is the normal to the mirror at the point of incidence P'.
For a concave mirror,
∠BP'C = ∠P'CF = θ ...[alternate angles]
and ∠BP'C = ∠CP'F = θ ...[law of reflection, ∠i = ∠r]
Hence ∠P'CF = ∠CP'F
∴ ΔFP'C is isosceles.
Hence, P'F = FC
If the aperture of the mirror is small, the point P' is very close to the point P, then:
P'F = PF
∴ PF = FC
= `1/2 PC`
∴ f = `1/2 R`
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