Physics

Problem

(a) Starting from appropriate similar triangles in the ray diagram for a concave spherical mirror, derive the mirror equation

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

. In your derivation, clearly show where the condition

f = \dfrac{R}{2}

is used. (b) Using the result obtained in part (a), prove that when the object is placed at the centre of curvature

C

(i.e.,

u = -R

), the image is formed at the same point

C

(that is,

v = -R

), and explain the nature and magnification of this image.

Intuitive explanation

PART (A)

The mirror equation emerges from the geometry of ray paths—specifically, two similar triangles formed by the object, image, and mirror's optical axis. By writing the ratios of corresponding sides from these triangles and using the relationship between focal length and radius of curvature (

f = R/2

), you'll eliminate the object and image heights to arrive at the reciprocal form. The key is recognizing that this geometric constraint, combined with the paraxial approximation, forces a unique relationship between object distance

u

, image distance

v

, and focal length

f

PART (B)

Once you have the mirror equation, substituting

u = -R

is straightforward algebra—the equation will yield

v = -R

directly. The physical insight is that the center of curvature is special: rays from an object there reflect back through the same point, making the image real, inverted, and equal in size to the object (magnification

m = 1

). This is the only position where object and image coincide.