Chapter 23, Problem 52P

To determine

(a) To Determine:

Ray diagram and describe the type of image formed.

Answer to Problem 52P

Solution:

Explanation of Solution

Given:

Book is placed $9.0cm$ behind a lens it is desired to magnify reading material by a factor $3.0x$ i.e., Object distance, $u=-9cm$. Image distance, $v=27cm$

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:

Image location 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., $u=-9cm$ and $v=27cm$. Thus substituting them in formula we get-

$\begin{array}{l}-\frac{1}{27}+\frac{1}{9}=\frac{1}{f}\\ \frac{1}{f}=\frac{2}{27}\\ f=13.5cm\\ P=\frac{1000}{f}\\ P=\frac{1000}{135}\\ P=7.41\end{array}$[where $f$ in mm]

Virtual and magnified image is formed.

To determine

(b) To Determine:

Type of the lens needed.

Answer to Problem 52P

Solution:

Converging lens needed.

Explanation of Solution

Given:

Book is places $9.0cm$ behind a lens it is desired to magnify reading material by a factor $3.0x$ i.e., Object distance, $u=-9cm$. Image distance, $v=27cm$

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:

Converging is used.