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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Experimental Physics

NTA Abhyas 2020 · Physics · Experimental Physics

Graph of position of image vs position of a point object from a...

Graph of position of image vs position of a point object from a convex lens is shown in the figure. The focal length of the lens is –

& Graph of position of image vs position of a point object from a convex lens is shown in the figure. The focal length of

Options

  1. A(0.50±0.05)cm\left(\text{0.50} ± \text{0.05}\right) \text{cm}
  2. B(5.00±0.05)cm\left(\text{5.00} ± \text{0.05}\right) \text{cm}
  3. C(0.50±0.10)cm\left(\text{0.50} ± \text{0.10}\right) \text{cm}
  4. D(5.00±0.10)cm\left(\text{5.00} ± \text{0.10}\right) \text{cm}

Correct answer

B. (5.00±0.05)cm\left(\text{5.00} ± \text{0.05}\right) \text{cm}

Step-by-step solution

2f=102 \text{f} = 1 0

f=5cm\text{f} = 5 cm

1f=1v−1u\dfrac{1}{\text{f}} = \dfrac{1}{\text{v}} - \dfrac{1}{\text{u}}

dff2=dvv2+duu2\dfrac{\text{df}}{f^{2}} = \dfrac{\text{dv}}{v^{2}} + \dfrac{\text{du}}{u^{2}}

df=f2(dvv2+duu2)<br/><br/>u=10,v=10, and du=110,dv=110<br/><br/>(asthereare10divisionin1cmscale,leastcountis0.1i.e.,consideredaserror)<br/><br/>Sodf=25(110(10)2+110(10)2)=25[210×100]<br/><br/>df=0.05sof=(5.00±0.05)cm.\text{df} = f^{2} \left(\dfrac{\text{dv}}{v^{2}} + \dfrac{\text{du}}{u^{2}}\right)<br /><br />u = 10, v = 10, \text{ and } \text{du} = \dfrac{1}{1 0}, \text{dv} = \dfrac{1}{1 0}<br /><br />(as there are 10 division in 1 cm scale, least count is 0.1 i.e., considered as error)<br /><br />So \text{df} =2 5 \left(\dfrac{1}{1 0 {\left(1 0\right)}^{2}} + \dfrac{1}{1 0 { \left(1 0\right)}^{2}}\right) =2 5 \left[\dfrac{2}{1 0 × 1 0 0}\right]<br /><br />df = 0.05 so f =(5.00 ± 0.05) cm.
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