Animated Solution for Physics - Optics: The graph between object distance $u$ and image distance $v$ for a lens is given below. The focal length of the lens is

When you first look at the problem, you are presented with a graph plotting the image distance $v$ against the object distance $u$ for a lens. The key to solving this problem lies in extracting the right data points from this visual representation.

Notice the specific point where the curve intersects the $45^{\circ }$ line. At this exact intersection, the object distance $u$ is $-10\text{ cm}$, and the image distance $v$ is $+10\text{ cm}$.

But that's not all! We also need to determine the precision of our measurements. If you observe the grid closely, there are $10$ subdivisions between the markings $9$ and $10$. This means the least count, or the maximum possible error in measuring $u$ and $v$, is $\Delta u=0.1\text{ cm}$ and $\Delta v=0.1\text{ cm}$. This tiny detail is the real catch of the problem!

Now, to find the focal length of the lens, we will use the standard lens formula:

Let's substitute the values we extracted from the graph into our formula. We plug in $v=+10$ and $u=-10$. Be very careful with the minus sign here, as it's a common place for silly mistakes.

Simplifying the expression, the two negative signs become positive, giving us:

Taking the reciprocal, we find that the focal length $f$ is exactly $+5\text{ cm}$. The positive sign confirms that it's a convex lens.

We have the focal length, but the options also include the error limits. To find the error in the focal length, $\Delta f$, we need to differentiate the lens formula.

Differentiating $\frac{1}{f}$ gives $-\frac{\Delta f}{f^2}$. Applying this to the whole equation and taking the maximum permissible error by adding the absolute values, we get:

Now, let's rearrange the equation to solve for $\Delta f$. We multiply the entire right side by $f^2$:

Next, we substitute all our known values: $f=5$, $v=10$, $u=10$, and both $\Delta v$ and $\Delta u$ are $0.1$:

Let's calculate the final error. Inside the bracket, we have $\frac{0.1}{100}+\frac{0.1}{100}$, which adds up to $\frac{0.2}{100}$.

Multiplying this by $25$ (the square of $5$), we get:

Finally, we combine our results. The focal length of the lens, along with its maximum permissible error, is written as $f\pm \Delta f$. This gives us $5\pm 0.05\text{ cm}$. This matches perfectly with the correct option!