Chapter 23, Problem 23P

To determine

Part (a) To Determine:

The magnification by two following techniques.

By ray diagram; (i)$\left(d_0>r\right),m<1$ (ii)$\left(d_0<r\right),m>1$

Answer to Problem 23P

Solution:

Since $\left(d_0>r\right),$ then $\frac{r}{\left(2d_0-r\right)>r}orm<1$

Explanation of Solution

Given Info:

Let height of object be $+h_0$

Height of image $h_i$

Object distance $d_0$

Image distance $d_i$

Let the focal length and radius of curvature of mirror be ‘f’ and ‘r’ respectively.

Formula Used:

Magnification of mirror $m=\frac{h_i}{h_0}=-\frac{d_i}{d_0}$

Here, all alphabets are in their usual meanings.

Calculation:

The magnification by ray diagram.

(i)$\left(d_0>r\right),m<1$

(ii)$\left(d_0<r\right),m>1$

To determine

Part (b) To Determine:

The magnification by two following techniques.

By mirror equation (i)$\left(d_0>r\right),m<1$ (ii)$\left(d_0<r\right),m>1$

Answer to Problem 23P

Solution:

Since $\left(d_0<r\right),$ then $\frac{r}{\left(2d_0-r\right)<r}orm>1$

Explanation of Solution

Given Info:

Let height of object be $+h_0$

Height of image $h_i$

Object distance $d_0$

Image distance $d_i$

Let the focal length and radius of curvature of mirror be ‘f’ and ‘r’ respectively.

Formula Used:

(i)Apply focal length formula for mirror

$\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_0}$

(ii)Magnification of mirror $m=\frac{h_i}{h_0}=-\frac{d_i}{d_0}$

Here, all alphabets are in their usual meanings.

Calculation:

The magnification by using mirror equation:

First, we shall find the image distance ‘d_i’

Apply focal length formula for mirror

$\begin{array}{l}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_0}\\ \frac{2}{r}=\frac{1}{d_i}+\frac{1}{d_0}\\ or,d_i=\frac{d_0\times r}{\left(2d_0-r\right)}\end{array}$

Apply magnification of mirror $m=-\frac{d_i}{d_0}=-\frac{r}{\left(2d_0-r\right)}-----(i)$

Since $\left(d_0>r\right),$ then from equation (i)$\frac{r}{\left(2d_0-r\right)>r}orm<1$

Again, Since $\left(d_0<r\right),$ then from equation (i)$\frac{r}{\left(2d_0-r\right)<r}orm>1$