Question

Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the centre of curvature. Using this diagram, derive the mirror formula.

Hint:

For a concave mirror: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] Magnification: \[ m=-\frac{v}{u} \] Object between \(F\) and \(C\):
• Image beyond \(C\)
• Real
• Inverted
• Enlarged

Answer 1

Concept: A concave mirror is a spherical mirror whose reflecting surface is curved inward. When an object is placed between the focus \(F\) and the centre of curvature \(C\), the image formed is:
• Real
• Inverted
• Magnified
• Formed beyond the centre of curvature The mirror formula establishes a relationship among: \[ u=\text{object distance}, \] \[ v=\text{image distance}, \] and \[ f=\text{focal length}. \] The required relation is \[ \boxed{\frac{1}{f}=\frac{1}{v}+\frac{1}{u}} \] which is valid for all spherical mirrors when sign convention is properly followed.

Step 1: Draw the ray diagram. Consider a concave mirror with pole \(P\), focus \(F\), and centre of curvature \(C\). The object \(AB\) is placed between \(F\) and \(C\). \[ \text{Ray 1: Parallel to principal axis } \longrightarrow \text{ reflected through } F \] \[ \text{Ray 2: Through } C \longrightarrow \text{ reflected back along the same path} \] The reflected rays intersect beyond \(C\) at point \(A'B'\), forming a real and inverted image. Schematic Ray Diagram

Step 2: Introduce geometrical quantities. Let \[ PB=u \] be the object distance, \[ PB'=v \] be the image distance, and \[ PF=f \] be the focal length. Let the object height be \[ AB=h \] and image height be \[ A'B'=h'. \]

Step 3: Use similar triangles. From the geometry of the ray diagram, \[ \triangle ABP \sim \triangle A'B'P. \] Therefore, \[ \frac{AB}{A'B'} = \frac{PB}{PB'}. \] Hence, \[ \frac{h}{h'} = \frac{u}{v}. \] Thus, \[ \boxed{ \frac{h'}{h} = \frac{v}{u} } \] which is the expression for magnification.

Step 4: Consider triangles involving the focus. From the ray travelling parallel to the principal axis and then passing through the focus after reflection, the triangles formed in the geometry of reflection are similar. Using the similarity of the appropriate triangles obtained from the ray diagram, \[ \frac{AB}{A'B'} = \frac{FP}{F B'}. \] Substituting distances, \[ \frac{h}{h'} = \frac{f}{v-f}. \] Using \[ \frac{h}{h'} = \frac{u}{v}, \] we obtain \[ \frac{u}{v} = \frac{f}{v-f}. \] Cross-multiplying, \[ u(v-f)=vf. \] Expanding, \[ uv-uf=vf. \] Rearranging, \[ uv=uf+vf. \] Dividing throughout by \(uvf\), \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v}. \] Hence, \[ \boxed{ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} } \] This is the required mirror formula. Final Result: \[ \boxed{ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} } \]