Table of Contents
For Concave Mirror
Let the concave spherical mirror has focal length f and radius of curvature R. Let P, F and C be its pole, focus and centre of curvature respectively. Consider a ray of light OA is incident on the mirror which is parallel to the principal axis of the mirror. Join AC. Then, AC is normal to the mirror at point A. Let the ray of light gets reflected along AB cutting the pricipal axis at point F. Let i be the angle of incidence and r be the angle of reflection.
Then, \[\text{angle of incidence} = \text{angle of reflection}\] \[∠OAC=∠CAF\text{ __(1)}\] also, \[∠OAC= ∠ACF\text{ __(2) [Being alternate angles]}\] From (1) and (2), \[∠CAF=∠ACF\] Thus, ΔACF is an isosceles triangle. \[∴CF=AF\text{ __(3)}\] Now, \[CP=CF+FP\text{ __(4)}\] From (3) and (4), \[CP=AF+FP\text{ __(5)}\] The points A and P lies very close to each other because the aperture of the mirror is small. Thus, \[\text{when }A→P\text{,}\] \[AF ≈FP\] Therefore, equation (4) becomes, \[CP=FP+FP\] \[CP=2FP\] \[FP=\frac{1}{2}CP\] \[∴f=\frac{1}{2}R\] This is the relation between focal length and radius of curvature for concave mirror.
For Convex Mirror
In the similar way, let the convex spherical mirror has focal length f and radius of curvature R. Let P, F and C be its pole, focus and centre of curvature respectively. Consider a ray of light OA is incident on the mirror which is parallel to the principal axis of the mirror. Then, the ray of light gets reflected along AB. Join AC and produce outward. Then, AC is normal to the mirror at point A. Let i be the angle of incidence and r be the angle of reflection.
Then, \[\text{angle of incidence} = \text{angle of reflection}\] \[∠OAN= ∠BAN\text{ __(a)}\] Now, \[∠FAC=∠BAN\text{ __(b) [Being vertically opp. angles]}\] \[∠FCA=∠OAN\text{ __(c) [Being corresponding angles]}\] From (a),(b) and (c), \[∠FAC=∠FCA\] Thus, ΔACF is an isosceles triangle. \[∴CF=AF\text{ __(3)}\] Now, \[CP=CF+FP\text{ __(4)}\] From (3) and (4), \[CP=AF+FP\text{ __(5)}\] Since the aperture of the mirror is small, \[\text{when }A→P\text{,}\] \[AF ≈FP\] Therefore, equation (4) becomes, \[CP=FP+FP\] \[CP=2FP\] \[FP=\frac{1}{2}CP\] \[∴f=\frac{1}{2}R\] This is the relation between focal length and radius of curvature for convex mirror.