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Derive mirror formula of convex mirror​

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Mirror formula

Let the object distance (u), image distance (v) and focal length (f). 

The following sign convention is used for measuring various distances in the ray diagrams of spherical mirrors:

  • All distances are measured from the pole of the mirror.
  • Distances measured in the direction of the incident ray are positive and the distances measured in the direction opposite to that of the incident rays are negative.
  • Distances measured above the principal axis are positive and that measured below the principal axis are negative.

 

Similarly, In \triangle{FPE} and  

\begin{array}{l}{\frac{E P}{A^{\prime} B^{\prime}}=\frac{P F}{A^{\prime} F}} \\ \\ {\frac{A B}{A^{\prime} B^{\prime}}=\frac{P F}{A^{\prime} F}[\mathrm{AB}=\mathrm{EP}] \cdots(\mathrm{II})}\end{array}

From (i) &(ii) 

\begin{array}{l}{\frac{A C}{A^{\prime} C}=\frac{P F}{A^{\prime} F}} \\ \\ {=>\frac{A^{\prime} C}{A C}=\frac{A^{\prime} F}{P F}} \\ \\ {=>\frac{\left(C P-A^{\prime} P\right)}{(A P-C P)}=\frac{\left(A^{\prime} P-P F\right)}{PF} }\end{array}

 

Now, PF = -f  ;  CP=2PF=-2f ; AP= -uAP=-u \ \text{and} \ A'P=-v

Put these value in above relation: 

\begin{array}{l}{\Longrightarrow \frac{[(-2 f)-(-v)]}{(-u)-(-2 f)}=\frac{[(-v)-(-f)]}{(-f)}} \\ \\ {\Longrightarrow u v=f v+u f} \\ \\ {\Longrightarrow \frac{1}{f}=\frac{1}{u}+\frac{1}{v}}\end{array}

Posted by

shubham.krishnan

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