Question

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 1

$\boxed{\bf{ \blue{\bigstar Find \longrightarrow }}}$

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)

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Extra info-

• Nature of image – Real and inverted....
• for figure refer the attachment

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Answer 2

By sign convention ,

Object height (Ho) = 5 cm

Focal length (f) = 10 cm

Object distance (u) = -25 cm

We know that , the lens formula is given by

$\boxed{ \sf \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }$

Thus ,

1/10 = 1/v - (-1/25)

1/v = 1/10 - 1/25

1/v = (25 - 10)/250

v = 250/15

v = 16.6 cm

★ The distance of image is 16.6 cm from the lens on the right side

Now , the magnification of lens is given by

$\boxed{ \sf{m = \frac{ h_{i}}{h_{o}} = \frac{v}{u} }}$

Thus ,

hi/5 = -16.6/25

hi = -16.6/5

hi = -3.3 cm

The negative sign of hi shows the image is inverted

★ The height of object is 3.3 cm