# Sir, how to prove R=2f and mirror formalae i.e 1/f=1/u+1/v in spherical mirror

### Asked by Priyansh Agrawal | 30th Oct, 2014, 08:46: PM

** To prove R=2f: **
Consider a ray of light AB, parallel to the principal axis, 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 mirror will pass through F (concave mirror) or will appear to diverge from F (convex mirror) and obeys law of reflection, i.e., i = r.
From the geometry of the figure, if the aperture of the mirror is small, B lies close to P.
Therefore, BF = PF
or FC = FP = PF
or PC = PF + FC
= PF + PF
or R = 2PF
or R = 2f
**Mirror formalae in spherical mirror:**

In the figure shown above, an object AB is placed at a distance u from the pole of the concave mirror of small aperture, just beyond the centre of curvature. Hence, its real, inverted and diminished image A’B’ is formed at a distance v in front of the mirror.

A/C to Cartesian sign convention,

Object distance (PB) = -u

Image distance (PB’) = -v

Focal length (PF) = -f

Radius of curvature (PC) = -R

It is clear from the geometry of figure, right angle ABP and A’B’P’ are similar.

**To prove R=2f:**

**Mirror formalae in spherical mirror:**

In the figure shown above, an object AB is placed at a distance u from the pole of the concave mirror of small aperture, just beyond the centre of curvature. Hence, its real, inverted and diminished image A’B’ is formed at a distance v in front of the mirror.

A/C to Cartesian sign convention,

Object distance (PB) = -u

Image distance (PB’) = -v

Focal length (PF) = -f

Radius of curvature (PC) = -R

It is clear from the geometry of figure, right angle ABP and A’B’P’ are similar.

### Answered by Yashvanti Jain | 8th Dec, 2017, 07:04: PM

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