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NEET
Exam
NEET
Subject
Physics
Chapter
Experimental Physics

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

Correct answer

B. (5.00±0.05)cm\left(\text{5.00} \pm \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}}{{\text{f}}^{2}} = \dfrac{\text{dv}}{{\text{v}}^{2}} + \dfrac{\text{du}}{{\text{u}}^{2}}

df=f2(dvv2+duu2)\text{df} = {\text{f}}^{2} \left(\dfrac{\text{dv}}{{\text{v}}^{2}} + \dfrac{\text{du}}{{\text{u}}^{2}}\right)

u = 10, v = 10,  du=110\text{du} = \dfrac{1}{1 0}, dv=110\text{dv} = \dfrac{1}{1 0}

(as there are 10 division in 1 cm scale, least count is 0.1 i.e., considered as error)

So df=25(110(10)2+110(10)2)=25[210×100]\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 \times 1 0 0}\right]

df = 0.05 so f =(5.00 ± 0.05) cm.
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