Question

Prove that focal length is half the radius of curvature

Answer 1

Focal length and radius of curvature:

• Let F be focus and C be center of curvature of a concave mirror.
• Consider an incident ray travelling parallel to principle axis and meet at point N.
• It passes through focus after reflection.
• CN is the normal drawn to the mirror(normals drawn to mirror meet at C).
• Now,

                   PC = R

                   PF = f

                   ∠FNC = ∠FCN = $\alpha$

                  ∠MFN = 2$\alpha$

• In ΔMNC

                 $tan\alpha = \frac{MN}{CM}$

                 $tan\alpha = \frac{MN}{PC}$                  [For paraxial rays, M and P coincide]

• If $\alpha$ is small, $tan\alpha = \alpha$

                  $\alpha = \frac{MN}{PC}$       ------------------------(1)

• In ΔMNF

                 $tan2\alpha = \frac{MN}{FM}$

                 $tan2\alpha = \frac{MN}{PF}$                  [For paraxial rays, M and P coincide]

• If $\alpha$ is small, $tan\alpha = \alpha$

                 $2\alpha = \frac{MN}{PF}$           ------------------------(2)

• Substituting (1) in (2), we get

                 $2 \frac{MN}{PC} = \frac{MN}{PF}$

                    $\frac{2}{PC} = \frac{1}{PF}$

                   2PF = PC

                       R = 2f

                        f = R/2

• Therefore, focal length is half the radius of curvature.

In fig, theta is taken instead of alpha.

 

Answer 2

Proving the focal length is half the radius of curvature:

Taking a concave mirror, the curved mirror will have a principal axis near which a ray of light is incident on the mirror parellel to it.

Now the angle between the radius of curvature and principal axis will be equal to the angle at which ray is incident and due to reflections law incident angle would be equal to angle cbf.

                                L abc = L cbf (law of reflection)

                                      L abc = L fcb ( alternate)

Hence, cbf = fcb ⇒ bf = fc

Since b very close to p; pf = fc ⇒ pc = 2 pf

Now considering convex mirror,  similar to concave mirror L abn = L nbd {reflection law}

L abn = Lbcf {corresponding}

Hence, bf = fc

Since, b is very close to p, pf = fc ⇒ pc = 2pf

Hence, in both cases Radius is double the focal length.

To know more:

Prove that focal length of the concave mirror is half of the radius of curvature

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