Question

An object 5 cm in length is held 25 cm away from a converging lens of focal length
10 cm. Draw the ray diagram and find the position, size and the nature of the
Image formed.​

Answer 1

Answer:

h1=5cm, u= -25cm f=10cm

Explanation:

by lens formula

1/v-1/u=1/f

then

1/v=1/f-1/u

1/v=1/10+1/-25

1/v=(5-2) /50

1/v=3/50

v=50/3 cm or16. 67cm

m=v/u

m=50/3/-25

m=-2/3

-2/3=h2/5

h2=-10/3cm or3. 33cm

real inverted and diminished

Answer 2

Given: An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm.

To Find: The position, size and the nature of theImage formed.

$\: \: \: \: \:$________________________

✞USING FORMULA:

$\:\:\: \: \: \: \: \: \: \color{blue}{ \underline{ \boxed{ \bf{ Mirror \: formula=\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} }}}} \: \red{\star }$

$\: \: \: \:: \implies{ \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15} }}</p><p> \\ \\ \\ \: \: \: \::\implies{ \sf{\dfrac{1}{v} = \dfrac{3}{50} }} \\ \\ \\</p><p> \: \: \: \:: \implies{ \sf \: v = \dfrac{50}{3} }</p><p>$

Hence, The image is real and inverted

✞ MAGNIFICATION:

$\: \: \: \: \: \: \: \:\: \: \: \: \color{blue}{ \underline {\boxed{ \bf{Magnification = \dfrac{ h_{1} }{ h_{o} } = \dfrac{v}{u} }}}} \: \red{ \star}$

$\: \: \: \: :\implies \sf \dfrac{ h_{1} }{5} = \dfrac{ - 16.7}{25} \\ \\ \\\: \: \: \: :\implies \sf h_{1} = \dfrac{ - 10}{3} cm$

$\\ {{ \underline{ \boxed {\sf\therefore The \: image \: is \: inverted \: and \: diminished.}}}}$

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