Question

(a) Show that if the object O in Fig. 34-19c is moved from focal point $F_1$toward the observer’s eye, the image moves in from infinity and the angle (and thus the angular magnification mu) increases. (b) If you continue this process, where is the image when mu has its maximum usable value? (You can then still increase, but the image will no longer be clear.) (c) Show that the maximum usable value of is$m_{\theta }=1+\frac{25cm}{f}$.(d) Show that in this situation the angular magnification is equal to the lateral magnification.

Short answer

1. Increase in the angular magnification as the object is moved from the focal point towards the observer’s eye is $\frac{d{m}_{\theta }}{dt}>0$$,\frac{dp}{dt}<0$
2. Position of image when $m_{\theta }$is maximum, $i=p_n$
3. Maximum usable value of localid="1664258675620" $m_{\theta }=1+\frac{25cm}{f}$
4. For maximum usable value of

$\begin{array}{c}{m}_{\theta }=\frac{h\text{'}}{h}\\ =\frac{\left|i\right|}{p_n}·m_{\theta }\end{array}$

Step-by-step solution

Listing the given quantities

The object is moved from the focal point towards the observer’s eye.

Near point of person’s eye is$p_n=25cm$

Understanding the concepts of angular magnification 

Angular magnification is a ratio of angular height with magnifier to the height of image without magnifier. Hence, when we differentiate it with respect to time, we get the relation in terms of object distance and the near vision of human eye. In a similar way, we can get the equation for change in image distance using differentiation of lens equation. By comparing the two equations, we can get the required result.

Formula:

Angular height$\theta =\frac{h}{p_n}$

Angular height with magnifier$\theta \text{'}=\frac{h\text{'}}{p}$

Angular magnification role="math" localid="1662987541762" $m_{\mathrm{\theta }}=\frac{\mathrm{\theta }\text{'}}{\mathrm{\theta }}$

Lens formula$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$

Explanation 

(a)

The position of the virtual image formed by the lens, which acts as an object for the objective of the lens:

$\frac{1}{p}=\frac{1}{f}-\frac{1}{i}$

As the image is virtual, $i=-\left|i\right|;$then using this, we get

$\frac{1}{p}=\frac{1}{f}+\frac{1}{\left|i\right|}$

Differentiating the above equation with respect to time $t$we get

$\begin{array}{l}\frac{d}{dt}\left(\frac{1}{p}\right)=\frac{d}{dt}\left(\frac{1}{f}\right)+\frac{d}{dt}\left(\frac{1}{\left|i\right|}\right)\\ -\frac{1}{p^2}\frac{dp}{dt}=-\frac{1}{{\left|i\right|}^2}\frac{d\left|i\right|}{dt}\\ \frac{1}{p^2}\frac{dp}{dt}=\frac{1}{{\left|i\right|}^2}\frac{d\left|i\right|}{dt}\\ \frac{dp}{dt}=\frac{p^2}{{\left|i\right|}^2}\frac{d\left|i\right|}{dt}\end{array}$

The object is moved from the focal point towards the observer’s eye; therefore, $\frac{dp}{dt}$is decreasing.

Hence$\frac{dp}{dt}<0,$which implies $\frac{d\left|i\right|}{dt}<0,$that is,theimage is moving in from infinity.

Now consider the angular magnification:

$\begin{array}{c}{m}_{\theta }=\frac{\theta \text{'}}{\theta }\\ =\left(\frac{\frac{h}{p}}{\frac{h}{p_n}}\right)\\ =\frac{p_n}{p}\end{array}$

Differentiating the above equation with respect to time $t$we get

$\begin{array}{c}\frac{d{m}_{\theta }}{dt}=p_n\frac{d}{dt}\left(\frac{1}{p}\right)\\ =\left(-\frac{p_n}{p^2}\right)·\frac{dp}{dt}\end{array}$

From the above equation, we clearly see that$\frac{d{m}_{\theta }}{dt}>0$

Angular magnification increases as the object is moved from focal point towards the observer’s eye.

Explanation of position of image when is maximum

(b)

Angular magnification,

$\begin{array}{c}{m}_{\theta }=\frac{\theta \text{'}}{\theta }\\ =\left(\frac{\frac{h}{p}}{\frac{h}{p_n}}\right)\\ =\frac{p_n}{p}\end{array}$

The image formed by the lens will appear to be near the point $\left|i\right|=p=p_n$. when the angular magnification is maximum because we know that the virtual image is formed by the lens, which acts as an object for the objective of the lens ($\left|i\right|=p$ ).

Explanation of maximum usable value ofmθ 

(c)

We have lens formula

$\frac{1}{p}=\frac{1}{f}-\frac{1}{i}$

Astheimage is virtual, then$i=-\left|i\right|;$then using this, we get

$\frac{1}{p}=\frac{1}{f}+\frac{1}{\left|i\right|}$

Rearranging the terms, we get

$p=\frac{\left|i\right|f}{\left|i\right|+f}$

For maximum angular magnification $\left|i\right|=p_n$

$p=\frac{p_{n}f}{p_n+f}$

$\begin{array}{c}{m}_{\theta }=\frac{p_n}{p}\\ =\frac{p_n}{\frac{p_{n}f}{p_n+f}}\\ =\frac{p_n+f}{f}\end{array}$

Maximum usable value is

$\begin{array}{c}{m}_{\theta }=1+\frac{p_n}{f}\\ =1+\frac{25cm}{f}\end{array}$

Maximum usable value of$m_{\theta }=1+\frac{25cm}{f}$

Explanation 

(d)

$\begin{array}{c}{m}_{\theta }=\frac{\theta \text{'}}{\theta }\\ =\frac{h\text{'}}{p}·\frac{p_n}{h}\\ =\frac{h\text{'}}{h}\frac{p_n}{p}\end{array}$

$m_{\theta }\approx \frac{h\text{'}}{h}·\frac{p_n}{\left|i\right|}$

Lateral magnification is $\frac{h\text{'}}{h}\approx m_{\theta }\frac{\left|i\right|}{p_n}$

When $\left|i\right|=p_{n,}$we get $\frac{h\text{'}}{h}=m_{\theta }.$This shows the equality in magnification.