Question

If for a spherical mirror object distance, $ u=(50.1\,\pm \,0.5)\,cm $ and image distance $ \upsilon =(20.1\,\pm \,0.2)\,cm, $ then focal length of the spherical mirror will be

• A. $ (14.3\,\pm \,0.1)\,cm $
• B. $ (14.3\,\pm \,0.5)\,cm $
• C. $ (30.1\,\pm \,0.1)\,cm $
• D. $ (25.3\,\pm \,0.5)\,cm $

Answer 1

Answer: (A)

We have $ =\frac{1}{f}=\frac{1}{u}+\frac{1}{v}=\frac{v+u}{uv} $
$ \therefore $ $ f=\frac{uv}{v+u} $
$ =\frac{(50.1)(20.1)}{(50.1+20.1)} $
$ =14.3\,cm $
From $ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} $
$ \frac{-\Delta t}{{{f}^{2}}}=\frac{-\Delta u}{{{u}^{2}}}+\frac{-\Delta v}{{{v}^{2}}} $
$ \Rightarrow $ $ \Delta f=\Delta u{{\left( \frac{f}{u} \right)}^{2}}+\Delta v{{\left( \frac{f}{u} \right)}^{2}} $
$ =0.5{{\left( \frac{14.3}{50.1} \right)}^{2}}+0.2{{\left( \frac{14.3}{20.1} \right)}^{2}} $
$ =0.04+0.10 $
$ =0.14\,cm $
$ \therefore $ $ f=(14.3\pm 0.1)\,cm $