91Ó°ÊÓ

Provided:

The thin lens formula is$\frac{1}{i}+\frac{1}{o}=\frac{1}{f}$

Where f is the focal length of the lens o and i are the distance from the lens to the object and image

$i=12;o=18$

This method is based on the derivatives of functions whose values must be calculated at certain locations, as the name implies.

Consider a function $\mathit{y}\mathbf{=}\mathit{f}\left(x\right)$, find the value of a function$\mathit{y}\mathbf{=}\mathit{f}\left(x\right)$when$\mathit{x}\mathbf{=}\mathit{x}\mathbf{\text{'}}$.

As an example, the derivative of a function$\mathit{y}\mathbf{=}\mathit{f}\left(x\right)$with respect toxwill be employed.

$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{=}\left(f\left(x\right)\right)$is Change in y with respect to change in x as$dx⇒0$.

If the value of $x=x\text{'}$from a value of x near it, such that the difference in the two values, dx , is vanishingly small, one can derive the change in the value of the function $y=f\left(x\right)$corresponding to the change dx in x . In practice, however, the concept of vanishingly small is not possible.

Differentials are utilized with the thin lens formula to establish a relationship that may be used to compute di , or the small variation in image distance, and then the new image distance $\left(i_n\right).df$on the other hand, is nearly equivalent to $∆f$. (for very large n ),

As a result, it is possible to rewrite the differences using differentials:

The thin lens formula is as follows:

$\frac{1}{i}+\frac{1}{o}=\frac{1}{f}$

On differentiate,

$\frac{-1}{i^2}di+\frac{-1^2}{o^2}do=0$

Then,

$di=\frac{-i^2}{o^2}do$

Substitute the provided values,

$$\begin{gathered} do=17.5-18 \\ =-0.5 \end{gathered}$$

Here,

$$\begin{gathered} di=\frac{-{\left(12\right)}^2}{{\left(18\right)}^2}×\left(-0.5\right) \\ =0.22 \end{gathered}$$

Therefore, the new can be calculated as:

$\begin{array}{l}{i}_n=i+di\\ =12+0.22\\ =12.22\end{array}$

Hence,$i_nâ‰\dots 12.2$