Chapter 23, Problem 25P

To determine

Part (a) To Determine:

The distance between concave mirror of given mirror.

Answer to Problem 25P

Solution:

(i)The distance between concave mirror $D=\left(1+\frac{1}{\sqrt{2}}\right)r$

Explanation of Solution

Given Info:

Let object distance be $u$ and image distance be $u$

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

Formula Used:

(i)Apply focal length formula for mirror

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

Here, all alphabets are in their usual meanings.

Calculation:

Consider the distance between wall (where image is formed)and concave mirror is ‘x’. Let ‘r’ be radius of curvature. Hence, the image distance ‘v’ will be ‘D’.

Since the object is placed exactly at the distance ‘r’ from the wall. So, object distance from the mirror will be ‘u’ = D – r

Now, apply focal length formula for mirror

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ or,\frac{1}{f}=\frac{1}{D}+\frac{1}{D-r}\\ or,\frac{1}{f}=\frac{1}{D}+\frac{1}{D-2f}\\ or,D^2-4fD+2f^2=0\\ or,D=\left(2\pm \sqrt{2}\right)f\end{array}$

Since the value D will never be less than 2f then we will consider only positive sign.

Hence, the distance between mirror and wall be

$\begin{array}{l}D=\left(2+\sqrt{2}\right)f\\ D=\left(2+\sqrt{2}\right)\frac{r}{2}\\ D=\left(1+\frac{1}{\sqrt{2}}\right)r\end{array}$

To determine

Part (b) To Determine:

The magnification of given mirror.

Answer to Problem 25P

Solution:

(ii)The magnification of concave mirror $m=-\left(\sqrt{2}+1\right)$

Explanation of Solution

Given Info:

Let object distance be $u$ and image distance be $u$

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

Formula Used:

(i)Apply focal length formula for mirror

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

(ii)Magnification of mirror $m=-\frac{v}{u}$

Here, all alphabets are in their usual meanings.

Calculation:

Consider the distance between wall (where image is formed)and concave mirror is ‘x’. Let ‘r’ be radius of curvature. Hence, the image distance ‘v’ will be ‘D’.

Since the object is placed exactly at the distance ‘r’ from the wall. So, object distance from the mirror will be ‘u’ = D – r

Now, apply focal length formula for mirror

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ or,\frac{1}{f}=\frac{1}{D}+\frac{1}{D-r}\\ or,\frac{1}{f}=\frac{1}{D}+\frac{1}{D-2f}\\ or,D^2-4fD+2f^2=0\\ or,D=\left(2\pm \sqrt{2}\right)f\end{array}$

Since the value D will never be less than 2f then we will consider only positive sign.

Hence, the distance between mirror and wall be

$\begin{array}{l}D=\left(2+\sqrt{2}\right)f\\ D=\left(2+\sqrt{2}\right)\frac{r}{2}\\ D=\left(1+\frac{1}{\sqrt{2}}\right)r\end{array}$

Apply magnification of mirror

$\begin{array}{l}m=-\frac{v}{u}\\ m=-\frac{D}{D-r}\\ m=-\frac{\left(1+\frac{1}{\sqrt{2}}\right)r}{\left(1+\frac{1}{\sqrt{2}}\right)r-r}\\ m=-\left(\sqrt{2}+1\right)\end{array}$