Chapter 23, Problem 43P

To determine

(a) To Determine:

Object distance using a ray diagram.

Answer to Problem 43P

Solution:

$u=-477.64\text{mm}$

Explanation of Solution

Given:

A sharp image is located $391mm$ behind a $215mm$ focal-length converging lens.

i.e. Image distance, $v=391mm$. Focal length, $f=215mm$.

Formula used:

Lens equation, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$u$ and $v$ are lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.

Magnification, $m=\left|\frac{v}{u}\right|$

Calculation:

From the equivalence of the triangle $1$ and $2$ we can write-

$\begin{array}{l}\frac{h}{f}=\frac{mh}{v-f}\\ \frac{1}{215}=\frac{m}{391-215}\\ \Rightarrow m=\frac{391-215}{215}=\frac{176}{215}\end{array}$

Magnification $m$ is given by-

$\begin{array}{l}m=\frac{v}{u}\\ u=\frac{391}{m}\\ =391\times \frac{215}{176}\\ =477.64mm\end{array}$

From the diagram we can see that the object is placed behind the lens. Thus, object distance is negative. So, using ray diagram we get $u$ as $-477.64mm$.

To determine

(b) To Determine:

Object distance by calculation.

Answer to Problem 43P

Solution:

$u=-477.64mm$

Explanation of Solution

Given:

A sharp image is located $391mm$ behind a $215mm$ focal-length converging lens.

i.e. Image distance, $v=391mm$. Focal length, $f=215mm$.

Formula used:

Lens equation, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$u$ and $v$ are lens to object and lens to image distances respectively, negative when opposite to optical path and positive when in the direction of optical path.

Magnification, $m=\left|\frac{v}{u}\right|$

Calculation:

Object distance can be calculated using the lens formula.

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

Here value of $f$ and $v$ are given, i.e., $f=391mm$ and $v=215mm$. Thus substituting them in formula we get-

$\begin{array}{l}\frac{1}{391}-\frac{1}{u}=\frac{1}{215}\\ \frac{1}{u}=\frac{1}{391}-\frac{1}{215}\\ u=-\frac{391\times 215}{391-215}\\ u=-477.64mm\end{array}$

Negative sign indicate the direction.

Thus, object distance $(u)$ from calculation is $-477.64mm$.