Question

Illustration 5 A thin rod of length f/3 is placed along the optical axis of a concave mirror of focal length f such that its image which is real and elongated just touches the rod. Calculate the magnification.

JEE/Engineering Exams

Physics

Solution

165

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Given that Rod, ( A B ) is placed along the optical axis. The image of the rod will touch the rod, ( A B ) only when the end Cof the rod is placed at the centre of curvature of the concave mirror. ( mathrm{PC}=2 mathrm{f} ) and ( mathrm{CA}=mathrm{f} / 3 ) The image of the end ( C ) of the rod will be formed at ( C ) itself in this situation At A side of the rod: ( u=P A=P C-A C=2 f-(f / 3)=5 f / 3 ) Using the mirror formula, we get: ( frac{1}{f}=frac{1}{v}+frac{1}{u} ) Or, ( frac{1}{v}=frac{1}{f}-frac{1}{u} ) ( =frac{1}{f}-frac{1}{(5 f / 3)} ) ( =frac{1}{f}-frac{3}{5 f}=frac{2}{5 f} ) Or, ( v=frac{5 f}{2} ) Thus, the image of A of the rod is formed at ( A^{prime} ) which is at a distance ( 5 f / 2 ) from the pole of the mirror. ( mathrm{So}, mathrm{PA}^{prime}=5 mathrm{f} / 2 ) Length of the image of the ( mathrm{rod}, mathrm{CA}^{prime}=mathrm{PA}^{prime}-mathrm{PC}=(5 mathrm{f} / 2)-2 mathrm{f}=mathrm{f} / 2 ) Magnification, ( M=frac{text { Size of image }}{text { Size of object }} ) Or, ( M=frac{C A^{prime}}{C A} ) ( M=frac{f / 2}{f / 3}=frac{3}{2}=1.5 )