91影视

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

Near point of person鈥檚 eye is$p_n=25cm$

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$胃=\frac{h}{p_n}$

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

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

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

(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鈥檚 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:

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

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

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

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

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

(b)

Angular magnification,

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

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 ($\left|i\right|=p$ ).

(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 $\left|i\right|=p_n$

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

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

Maximum usable value is

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

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

(d)

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

$m_{胃}鈮�\frac{h\text{'}}{h}路\frac{p_n}{\left|i\right|}$

Lateral magnification is $\frac{h\text{'}}{h}鈮�m_{胃}\frac{\left|i\right|}{p_n}$

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