Question

The graph shows relationship between object distance and image distance for an equiconvex lens. Then, focal length of the lens is

• A. $0.50\pm 0.05cm$
• B. $0.50\pm 0.10cm$
• C. $5.00\pm 0.05cm$
• D. $5.00\pm 0.10cm$

Answer 1

The correct option is C $5.00\pm 0.05cm$

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$....... Equation 1

Consider the point (-10,10)

We get, $\frac{1}{f}=\frac{1}{5}$

$f=5cm$

Also differentiating Equation 1 gives,

$\frac{\Delta f}{f^2}=\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}$

From the graph we see that the least counts, $\Delta u,\Delta v=0.1cm$

Thus $\Delta f=\left(5^2\right)(\frac{0.1}{{10}^2}+\frac{0.1}{{10}^2})=0.05cm$