Question

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

Where f is the focal length of the lens and oand iare the distances from the lens to the object and image. If $\mathbf{i}\mathbf{=}\mathbf{15}$ ,when $\mathit{o}\mathbf{=}\mathbf{10}$use differentials to find i whenlocalid="1659153107776" role="math" $\mathbf{o}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{1}$.

Short answer

The answer is $i_n\cong 14.8$.

Step-by-step solution

Explanation of Solution

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.

$\mathrm{i}=15;\mathrm{o}=10$

Approximation by differentials

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

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

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

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

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.

Calculation

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 ),

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,

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

Then,

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

Substitute the provided values,

$$\begin{gathered} d=10.1-10 \\ =0.1 \end{gathered}$$

Here,

$$\begin{gathered} di=\frac{-{\left(15\right)}^2}{{\left(10\right)}^2}\times 0.1 \\ =-0.225 \end{gathered}$$

Therefore, the new can be calculated as:

$\begin{array}{l}{i}_n=i+di\\ =15+\left(-0.225\right)\\ =14.775\end{array}$

Hence,$i_n\cong 14.8$