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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Units and Dimensions

NTA Abhyas 2020 · Physics · Units and Dimensions

The focal length of a mirror is given by (1)/(f) = (1)/(u) +...

The focal length of a mirror is given by 1f=1u+1v\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} where uu and vv represent object and image distances respectively. The maximum relative error in ff is

Options

  1. A∆ff=∆uu+∆vv\dfrac{∆ f}{f} = \dfrac{∆ u}{u} + \dfrac{∆ v}{v}
  2. B∆ff=∆uu+∆vv−∆(u+v)u+v\dfrac{∆ f}{f} = \dfrac{∆ u}{u} + \dfrac{∆ v}{v} - \dfrac{∆ \left(u + v\right)}{u + v}
  3. C∆ff=∆uu+∆vv−∆(u−v)u+v\dfrac{∆ f}{f} = \dfrac{∆ u}{u} + \dfrac{∆ v}{v} - \dfrac{∆ \left(u - v\right)}{u + v}
  4. D∆ff=∆uu+∆vv+∆uu+v+∆vu+v\dfrac{∆ f}{f} = \dfrac{∆ u}{u} + \dfrac{∆ v}{v} + \dfrac{∆ u}{u + v} + \dfrac{∆ v}{u + v}

Correct answer

D. ∆ff=∆uu+∆vv+∆uu+v+∆vu+v\dfrac{∆ f}{f} = \dfrac{∆ u}{u} + \dfrac{∆ v}{v} + \dfrac{∆ u}{u + v} + \dfrac{∆ v}{u + v}

Step-by-step solution

f = \dfrac{uv}{u + v}, \dfrac{∆ f}{f} = \dfrac{∆ u}{u} + \dfrac{∆ v}{v} + \dfrac{Δ( u + v )}{u + v}
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