an object is 5 cm in size is paced 25cm away from a coverging lens of

1.	an object is 5 cm in size is paced 25cm away from a coverging lens of focal length 10cm .then size of image is
Answer» $\boxed{\bf{ \blue{\bigstar Find \longrightarrow }}}$ size of image? $\boxed{\bf{ \orange{\bigstar Given \longrightarrow }}}$ Height of OBJECT = 5cm Position of object, u = - 25cm Focal length of the lens, f = 10 cm $\boxed{\bf{ \green{\bigstar formulas \ used \longrightarrow }}}$ $\sf{ m= \dfrac{h_i}{h_o} }$ $\sf{ magnification = \dfrac{v}{u}}$ $\sf{\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}$ where, m= magnification, $h_i$ = height of image, $h_o$ = height of object, v= POSTION of mirror, u= position of object, f= focal length. $\boxed{\bf{ \pink{\bigstar SoLution \longrightarrow }}}$ Firstly we using mirror formula to find position of mirror, $\sf{\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}$ put given value in the formula we get, $\sf{\dfrac{1}{v} + \dfrac{1}{25} = \dfrac{1}{10}}$ $\sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{25} }$ $\sf{\dfrac{1}{v} = \dfrac{(5 - 2)}{50} }$ $\sf{\dfrac{1}{v} = \dfrac{3}{50} }$ $\sf{ v= \dfrac{50}{3} = 16.66 cm}$ Now we use magnification formula for finding the value of m, $\sf{magnification = \dfrac{v}{u}}$ put value in the formula we get, $\sf{m = \dfrac{16.66}{-25} = -0.66}$ now put the value of m and ho in given formula, $\sf{ m= \dfrac{h_i}{h_o} }$ from this we will find the size of image, $\sf{-0.66 = \dfrac{h_i }{ 5} }$ $\boxed{\sf{ \star h_i= -3.3 cm \star }}$ so the height of image = size of image $\bf{\underline{\therefore size \ of \ image\ is\ -3.3 cm }}$ ( at the opposite side of lens) _______________________________ Extra info- Nature of image – Real and inverted.... for figure refer the attachment _______________________________