Optics 3 Question 15
15. The graph between object distance $u$ and image distance $v$ for a lens is given below. The focal length of the lens is
(2006, 3M)
(a) $5 \pm 0.1$
(b) $5 \pm 0.05$
(c) $0.5 \pm 0.1$
(d) $0.5 \pm 0.05$
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Answer:
Correct Answer: 15. (b)
Solution:
- From the lens formula,
$$ \begin{array}{rlrl} & \frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \text { we have, } \\ & \frac{1}{f} & =\frac{1}{10}-\frac{1}{-10} \\ \text { or } & f u r t h e r, & =+5 \\ \text { and } \quad \Delta u & =0.1 \\ & \Delta v & =0.1 \quad \text { (from the graph) } \end{array} $$
Now, differentiating the lens formula, we have
$$ \text { or } \quad \begin{aligned} \frac{\Delta f}{f^{2}} & =\frac{\Delta v}{v^{2}}+\frac{\Delta u}{u^{2}} \\ \Delta f & =\frac{\Delta v}{v^{2}}+\frac{\Delta u}{u^{2}} f^{2} \end{aligned} $$
Substituting the values, we have
$$ \begin{aligned} \Delta f & =\frac{0.1}{10^{2}}+\frac{0.1}{10^{2}}(5)^{2}=0.05 \\ \therefore \quad f \pm \Delta f & =5 \pm 0.05 \end{aligned} $$
