Question

$\huge\fbox\red{❥Question}$

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{cccc}\sf \pink{Position_{\:(object)}} &\sf \purple{Position_{\:(image)}} &\sf \red{Size_{\:(image)}} &\sf \blue{Nature_{\:(image)}}\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf At \:Infinity &\sf At\: F&\sf Highly\:Diminished&\sf Real\:and\:Inverted\\\\\sf Beyond\:C &\sf Between\:F\:and\:C&\sf Diminished&\sf Real\:and\:Inverted\\\\\sf At\:C &\sf At\:C&\sf Same\:Size&\sf Real\:and\:Inverted\\\\\sf Between\:C\:and\:F&\sf Beyond\:C&\sf Enlarged&\sf Real\:and\;Inverted\\\\\sf At\:F&\sf At\:Infinity&\sf Highly\: Enlarged&\sf Real\:and\:Inverted\\\\\sf Between\:F\:and\:P&\sf \: Behind\:the\:mirror&\sf Enlarged&\sf \: Erect\:and\:Virtual\end{array}}\end{gathered}\end{gathered}\end{gathered}$
How far should an object be placed from the pole of a converging of focal length 20 cm to form a real image of the size exactly $\sf\dfrac{1}{4}$ th of the size of the object ?

Also, mention the case from the above table !

​

Answer 1

$\huge\fbox\red{❥Answer}$

f(focal length) = -20cm, m(size of image) = $-\dfrac{1}{4}$ (real image)

We know that

$\sf m = -\dfrac{v}{u}\\\sf -\dfrac{1}{4}= - \dfrac{v}{u}\\\sf u = 4v$

So,

$\sf\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\\\sf \dfrac{1}{v} + \dfrac{1}{4v} = \dfrac{1}{-20}\\\sf \dfrac{5}{4}v = - \dfrac{1}{20} v = -\dfrac{11}{4} = -25 cm$

∴ u = 4v= 4 × (-25) = - 100 cm

The object should be placed 100 cm to the left of the mirror.

The case here is second case from the table.

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{cccc}\sf \pink{Position_{\:(object)}} &\sf \purple{Position_{\:(image)}} &\sf \red{Size_{\:(image)}} &\sf \blue{Nature_{\:(image)}}\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf Beyond\:C &\sf Between\:F\:and\:C&\sf Diminished&\sf Real\:and\:Inverted\end{array}}\end{gathered}\end{gathered}\end{gathered}$

Hope it helps

Answer 2

$\large\underline{ \underline{ \sf \maltese{ \: Correct \: Question:- }}}$

How far should an object be placed from the pole of a converging mirror of focal length 20cm to form a real image of size exactly 1/4th of the size of object ?

$\large\underline{ \underline{ \sf \maltese{ \:Given:- }}}$

• f = -20cm
• Mirror= Converging/Concave

• $\sf{-hi=}{\dfrac{1}{4}}\times{ho}$

• Image formed is real and inverted.

$\large\underline{ \underline{ \sf \maltese{ \:To\:Find:- }}}$

• Distance of object from the pole of the mirror.

$\large\underline{ \underline{ \sf \maltese{ \:Solution:- }}}$

It is given that,

$\sf{-hi=}{\dfrac{1}{4}}\times{ho}$

Solving we get,

$\sf{{\dfrac{h_i}{h_o}}=-{\dfrac{1}{4}}=-{\dfrac{v}{u}}}$

$\sf{\implies}{v=}{\dfrac{u}{4}}$

From mirror formula,

$\sf{{\dfrac{1}{f}}={\dfrac{1}{v}}+{\dfrac{1}{u}}}$

Putting values in the formula:

$\sf{-{\dfrac{1}{20}}={\dfrac{4}{u}}+{\dfrac{1}{u}}}$

$\sf{\implies}{-{\dfrac{1}{20}}={\dfrac{1+4}{u}}}$

$\sf{\implies}{-{\dfrac{1}{20}}={\dfrac{5}{u}}}$

$\sf{\implies}{u=-100cm}$

Thus, object is placed at 100cm from the pole of the converging mirror.

Case: Case 2 from the table.

$\boxed{\begin{array}{c|c|c|c}\sf \pink{Position_{(object)}} &\sf \purple{Position_{(image)}} &\sf \red{Size_{(image)}} &\sf \blue{Nature_{(image)}}\\{\frac{\qquad \qquad \qquad \qquad}{}}&{\frac{\qquad \qquad \qquad \qquad}{}}&{\frac{\qquad \qquad \qquad \qquad}{}}&{\frac{\qquad \qquad \qquad \qquad}{}}\\\sf {Beyond\:C} &\sf {Between\:F\:and\:C}&\sf {Diminished}&\sf {Real\:and\:Inverted}\end{array}}$