CBSE Class 12 Physics 2020 Outside Delhi Set 3 Paper

Question : 12 of 13

Marks: +1, -0

A concave mirror forms a real image of an object kept at a distance

9 {\cm}

from it. If the object is taken away from the mirror by

6 {\cm}

, the image size reduces to

\;{1}/{4}

th of its previous size. Find the focal length of the mirror.

Solution:

For the

1^{\;\text "st "\;}

position :

\;\text " Object distance "\;\;=u_1=-9 {\cm}

\;\text " Image size "\;\;=h_1^{'}

\;\text " Object size "\;\;=h_1

Using mirror formula

\;{1}/{u}+\;{1}/{v}=\;{1}/{f}

\;\text " Or, "\;-\;{1}/{9}+\;{1}/{v_1}=\;{1}/{f}

.........(i)

\;\text " Magnification "\;=\;{h_1^{'}}/{h_1}=\;{v_1}/{u_1}

For the

2^{\;\text "nd "\;}

position :

\;\text " Object distance "\;\;=u_2=-15 {\cm}

\;\text " Image size "\;\;=h_2^{'}=\;{1}/{4} h_1^{'}

\;\text " Object size "\;\;=h_1

Using mirror formula

\;{1}/{u}+\;{1}/{v}=\;{1}/{f}

\;\;\text " Or, "\; \;\; -\;{1}/{15}+\;{1}/{v_2}=\;{1}/{f}

........(ii)

\;\;\text " Magnification "\;=\;{h_2^{'}}/{h_1}=\;{\;{1}/{4} h_1^{'}}/{h_1}=\;{v_2}/{u_2}

\;\;\text " Or, "\; \;\; \;{1}/{4} \;{v_1}/{u_1}=\;{v_2}/{u_2}

\;\;\text " Or, "\; \;\; v_2=\;{5}/{12} v_1

\;\;\text " Comparing equations (i) and (ii), "\;

\;-\;{1}/{9}+\;{1}/{v_1}=-\;{1}/{15}+\;{1}/{v_2}

\;\;\text " Or, "\; \;\; \;{1}/{15}-\;{1}/{9}=\;{1}/{v_2}-\;{1}/{v_1}

\;\;\text " Or, "\; \;\; -\;{2}/{45}=\;{12}/{5 v_1}-\;{1}/{v_1}

\;\;\text " Or, "\; \;\; -\;{2}/{45}=\;{7}/{5 v_1}

\; ∴ v_1=-\;{63}/{2} {\cm}

Putting in eqn (i),

\; \;\; -\;{1}/{9}-\;{2}/{63}=\;{1}/{f}

\;\;\text " Or, "\; \;\; -\;{1}/{7}=\;{1}/{f}

\; ∴ {\f}=\;\text " focal length "\;=-7 {\cm}

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