Question

The graph shows the relationship between object distance and image distance for an equi-convex 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

Hint: For a convex lens, the object distance and image distance will be the same, at 2f position. Lens formula can be used to find the equation of error in the focal length. We are also going to do this problem just like the least count problems.
Formula used:
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$, where f is the focal length, v is the image distance, and u is the object distance.

Complete step by step answer:
This graph shows the relation between the object distance and the image distance of a convex lens. When the object is moved closer to the lens, then the image distance also decreases. At the point of 2f, the image distance and object distance will be the same. For the further decreasing of object distance, the image distance will approach the infinity.

According to the graph,

$2f=10cm$
So focal length will be half of the 2f.

$f=\dfrac{10cm}{2}=5cm$

Next, we can find out the maximum error happening due to the measurement of focal length.

According to the lens formula,

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$, where f is the focal length, v is the image distance, and u is the object distance.

To find the maximum error, we can differentiate this equation.

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

Since we are calculating the error, no need to consider the negative sign. We can take the magnitude of these errors.

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

Here you can find $\Delta v$ from the divisions of v scale and $\Delta u$ from the divisions of u scale. Each centimetre has 10 divisions as shown in the figure.

$\Delta v=\dfrac{1}{10}=0.1$
$\Delta u=\dfrac{1}{10}=0.1$
We can assign these values to the equation.
$\dfrac{\Delta f}{{{5}^{2}}}=\dfrac{0.1}{{{10}^{2}}}+\dfrac{0.1}{{{10}^{2}}}$, here u and v are 10 cm since we are going to find the error in the focal length. It can be determined from that area only.

$\dfrac{\Delta f}{{{5}^{2}}}=\dfrac{0.2}{100}$
$\Delta f=0.05cm$

Therefore, the correct option is C and that is $5.00\pm 0.05cm$

Note: In these types of questions, we have to find out the least count. Least count is commonly used for the measurement of error of a system. Do not forget that, at 2f the image distance and object distance will be the same for the convex lens. That is the key point. Based on that we are measuring the error in focal length.