Question

5 cm high object is placed at a
distance of 25 cm from a converging
lens of focal length of 10 cm.
Determine the position, size and type
of the image.
(Ans : 16.7 cm, 3.3 cm, Real)​

Answer 1

$\huge{\boxed{\red{\mathcal{\underline{SOLUTION:}}}}}$

${\underline{\red{\bf{Given:}}}}\\ \\ \implies \sf Height\;of\;the\;object\;(H_{o})=5\;cm\\ \\ \implies \sf Object\;distance\;(u)=-25\;cm\\ \\ \implies \sf Focal\;length\;(f)=10\;cm\\ \\ \\ {\underline{\red{\bf{To\;Find:}}}}\\ \\ \implies \sf Position\\ \\ \implies \sf Size\\ \\ \implies \sf Type\;of\;the\;image\\ \\ \rule{200}{1}$

${\underline{\bf{Now,\;by\;using\;lens\;formula\;we\;will\;find\;image\;distance,}}}\\ \\ \\ \implies \sf \dfrac{1}{f} =\dfrac{1}{v} -\dfrac{1}{u}\\ \\ \\ \implies \sf \dfrac{1}{10}=\dfrac{1}{v}+\dfrac{1}{25}\\ \\ \\ \implies \sf \dfrac{1}{10}-\dfrac{1}{25} =\dfrac{1}{v}$

$\implies \sf \dfrac{5-2}{50}=\dfrac{1}{v}\\ \\ \\ \implies \sf \dfrac{3}{50}=\dfrac{1}{v}\\ \\ \\ \implies \sf 3v=50\\ \\ \\ \implies \sf v=\dfrac{50}{3}\\ \\ \\ \implies {\boxed{\green{\sf v=16.66\;cm}}}\\ \\ \rule{200}{1}$

${\underline{\bf Now,\;we\;will\;calculate\;Image\;height\;by\;magnification\;formula,}}\\ \\ \\ \implies \sf M=\dfrac{h_{i}}{h_{o}}=\dfrac{v}{u} \\ \\ \\ \implies \sf \dfrac{h_{i}}{5}=\dfrac{16.66}{25}\\ \\ \\ \implies \sf 25h_{i}=83.33\\ \\ \\ \implies \sf h=\dfrac{83.33}{25}\\ \\ \\ \implies {\boxed{\green{\sf h_{i}=3.33\;cm}}} \\ \\ \rule{200}{1}$

${\underline{\bf As\;image\;distance\;is\;positive.\;So,\;Real\;image\;is\;formed\;and\;image}}\\ \\ {\underline{\bf is\;formed\;on\;the\;opposite\;side\;of\;lens.}}\\ \\ {\underline{\bf As\;the\;height\;of\;the\;image\;is\;less\;than\;height\;of\;object.\;So,\;image}}\\ \\ {\underline{\bf is\;diminished.}}$

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

${\huge\tt\bf{Given}}\begin{cases}\sf{Height=5\:cm}\\\sf{Focal\:length=10\:cm}\\ \sf{u=-(25)\:cm}\end{cases}$

$\huge\tt\bf {To\:Find:-}$

• Position of image
• Size of image
• Type of image

$\huge\tt\bf {Solution:-}$

Finding the image distance :-

Lens formula is :-

$\huge\tt {\boxed {\dfrac {1}{v}-\dfrac {1}{u}=\dfrac {1}{f}}}$

$\leadsto\tt {\dfrac {1}{v}-\dfrac {1}{-(25)}=\dfrac {1}{10}}$

$\leadsto\tt {\dfrac {1}{v}=\dfrac {1}{10}-\dfrac {1}{25}}$

$\leadsto\tt {\dfrac {1}{v}=\dfrac {25-10}{250}}$

$\leadsto\tt {\dfrac {1}{v}=\dfrac {\cancel{15}}{\cancel {250}}}$

$\leadsto\tt {\dfrac {1}{v}=\dfrac {3}{50}}$

$\huge\tt {\red{\boxed {v=16.66\:cm}}}$

So, the position of the image is :- 16.66 cm.

$\rule{200}2$

Now, we know that :-

$\huge\tt {\boxed{\dfrac {{h}^{'}}{h}=\dfrac{v}{u}}}$

$\leadsto\tt {\dfrac {16.66}{-25}=\dfrac {{h}^{'}}{5}}$

$\leadsto\tt {{h}^{'}=\dfrac {16.66\times 5}{-25}}$

$\huge\tt {\red {\boxed {{h}^{'}=-3.33\:cm}}}$

So, the size of the image is -3.33 cm.

$\rule {200}2$

Now, as the size of the image is negative, we get to know that the image is real and inverted.

$\rule {200}2$

$\huge\tt\bf {Conclusion:-}$

• Position of image :- 16.66 cm
• Size of image :-3.33 cm
• Type of image :- Real and inverted

$\rule {200}5$