Chapter 25, Problem 23P

To determine

(a) To determine:

The position of the image.

Answer to Problem 23P

Solution:

$\text{2}.\text{46cm}$

Explanation of Solution

Given:

$\begin{array}{l}f=\text{5}.00\text{cm}\\ u=\text{4}.\text{85cm}\end{array}$

Formula used:

$\begin{array}{l}\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\ M=\frac{v}{u}\end{array}$

Calculation:

According to lens equation,

$\begin{array}{l}\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\ \frac{1}{v}=\frac{1}{f}-\frac{1}{u}\\ \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.0485}\\ \frac{1}{v}=40.61\\ v=2.46\text{cm}\end{array}$

To determine

(b) To determine:

The angular magnification:

Answer to Problem 23P

Solution:

$0.\text{5}0\text{72}X$

Explanation of Solution

Angular Magnification is given by,

$\begin{array}{l}M=\frac{\theta \text{'}}{\theta }=-\frac{v}{u}\\ M=-\frac{-0.0246}{0.0485}\\ M=0.5072\text{X}\end{array}$