Question

$\huge\red{\sf Heya!}$

An object 25cm high in front of a convex lens of focal length 30cm . If the height of image formed is 50 cm , find the distance between the object and the image ?

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Answer 1

$\huge\underline\green{\sf Answer :-}$

Distance between object = 135cm

Distance between image =15cm

$\huge\underline\green{\sf Solution:-}$

As object is in front of the lens , it is real and image is inverted and real so ,

$\sf{h_{1}}$ = 25 cm

f = 30 cm

$\sf {h_{2}=-50cm }$

$\large{\boxed{\sf m ={\frac{h_{2}{h_{1}}}}}}$

m = magnification of image .

$\large{\sf m ={\frac{-50}{25}}}$

$\large{\sf m = -2}$

$\large{\boxed{\sf m ={\frac{f}{f+u}}}}$

$\large{\sf -2={\frac{30}{30+u}}}$

$\large{\boxed{\sf u=-45}}$

$\large{\boxed{\sf m={\frac{v}{u}}}}$

$\large{\sf -2 ={\frac{v}{-45}}}$

$\large{\boxed{\sf v = 90cm}}$

As in the situation object and image are on opposite sides of lens , the distance between object and image

$\large{\sf d_{1} = u + v}$

$\large\implies{\sf 45+90}$

$\huge\red{\boxed{\sf d_{1}= 135cm }}$

If the image is erect (i.e virtual)

$\large{\boxed{\sf m ={\frac{f}{f+u}}}}$

$\large{\sf 2={\frac{30}{30+u}}}$

$\large{\boxed{\sf u = -15cm}}$

$\large{\sf m={\frac{v}{u}}}$

$\large{\sf 2={\frac{v}{15}}}$

$\large{\boxed{\sf v=30cm}}$

As in this situation both image and object are in front of the lens , the distance between object and image is

$\large{\sf d_{2}=v-u}$

$\large\implies{\sf 30-15}$

$\huge\red{\boxed{\sf d_{2} = 15cm}}$

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