Chapter 23, Problem 53P

To determine

(a) To Determine:

the distance by which the image is shifted if the object is moved $0.9m$ closer to the lens.

Answer to Problem 53P

Solution:

$3.54cm$

Explanation of Solution

Given:

Initially object is located $1.50m$ from a $6.5D$ lens i.e., Object distance, $u=-150cm$. Power, $P=6.5D$.

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

Power, $P=\frac{1000}{f}$

Where f is in mm

Calculation:

Given that $P=6.5D$, thus we can calculate focal length using the formula

$P=\frac{1000}{f}$

$\begin{array}{l}f=\frac{1000}{P}\\ f=\frac{1000}{6.5}\\ f=153.85mm\\ f=15.385cm\end{array}$

Thus, the position of the image formed can be calculated using Lens formula-

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

Here value of $u$ and $f$ are given, i.e., $u=-150cm$ and $f=15.385cm$. Thus substituting them in formula we get-

$\begin{array}{l}\frac{1}{v}+\frac{1}{150}=\frac{1}{15.385}\\ v=\frac{15.385\times 150}{150-15.385}\\ v=17.143cm\end{array}$

If the object is moved $0.9m$ closer to the lens,

$\begin{array}{l}u=-(150-90)\\ =-60cm\end{array}$

Thus, the position of the image formed after the shift can be calculated using Lens formula-

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

Here value of $u$ and $f$ are given, i.e., $u=-60cm$ and $f=15.385cm$. Thus substituting them in formula we get-

$\begin{array}{l}\frac{1}{v}+\frac{1}{60}=\frac{1}{15.385}\\ v=\frac{15.385\times 60}{60-15.385}\\ v=20.7cm\end{array}$

Change is $v=\left|v_{initial}-v_{final}\right|$

$\begin{array}{l}=20.7-17.143\\ =3.54cm\end{array}$

(Image moves further)

To determine

(b) To Determine:

the distance by which the image is shifted if the object is moved $0.9m$ farther from the lens.

Answer to Problem 53P

Solution:

$-0.7cm$

Explanation of Solution

Given:

Initially object is located $1.50m$ from a $6.5D$ lens i.e., Object distance, $u=-150cm$. Power, $P=6.5D$.

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

Power, $P=\frac{1000}{f}$

Where f is in mm

Calculation:

If the object is moved $0.9m$ farther from the lens,

$\begin{array}{l}u=-(150+90)\\ =-240cm\end{array}$

Thus, the position of the image formed after the shift can be calculated using Lens formula-

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

Here value of $u$ and $f$ are given, i.e., $u=-240cm$ and $f=15.385cm$. Thus substituting them in formula we get-

$\begin{array}{l}\frac{1}{v}+\frac{1}{240}=\frac{1}{15.385}\\ v=\frac{15.385\times 240}{240-15.385}\\ v=16.44cm\end{array}$

Change is $v=\left|v_{initial}-v_{final}\right|$

$\begin{array}{l}=16.44-17.143\\ =-0.7cm\end{array}$

(Image moves closer to the lens)