Question

An object of height 4.0 cm is placed at a distance 24 cm in front of a convex lens of focal length 8 cm.
★ Find the position and size of image. ​

Answer 1

Answer:

height of object =ho=4cm

distance of object = p=24cm

Focal lenght =f= 8cm

find

position of image = q =?

size of image = hi= ?

formula

1/q= 1/f-1/p

1/q=1/-8-1/24

1/q=-0.125-0.0416

1/q=-0.16666

q=-6.0024cm

formula

hi/ho=q/p

hi=-6.0024/24×4

hi=-1.000400cm

Answer 2

$\underline{ \large{ \sf{Information \: given \: to \: us:-}}}$

Height of object = 4.0 cm

Distance = 24 cm

Focal length = 8 cm

$\underline{ \large{ \sf{Need \: to \: calculate:-}}}$

Position of image

Size of image

$\underline{ \large{ \sf{Calculations:-}}}$

Now,

Lens formula:-

${\large: \longmapsto} \: \: \red{\boxed{\mathrm{ \dfrac{1}{v} - \dfrac{1}{u} = \frac{1}{f} }}}$

Where,

${ \large: \longmapsto} \: \mathrm{height \: is \: o}$

${ \large: \longmapsto} \: \: \mathrm{u \: is \: the \: distance \: of \: object}$

${\large: \longmapsto} \: \: \mathrm{focal \: length \: is \: f}$

By changing sides in formula,

${\large: \longmapsto} \: \: {\boxed{\mathrm{ \dfrac{1}{v} \: = \: \dfrac{1}{u} \: + \: \dfrac{1}{f} }}}$

Substituting values,

${\large: \longmapsto} \: \: {\mathrm{ \dfrac{1}{v} \: = \: \dfrac{1}{12} }}$

Again,

Using relationship,

${\large: \longmapsto} \: \: \red{\boxed{\mathrm{ \frac{ I}{O} \: = \: \frac{v}{u} }}}$

Where,

$: \large{ \longmapsto} \: \mathrm{I \: = \: image }$

$: \large{ \longmapsto} \: \mathrm{v\: = \: distance \: of \: image }$

$: \large{ \longmapsto} \: \mathrm{u\: = \: distance \: of \: object }$

Substituting values,

${\large: \longmapsto} \: \: {{\mathrm{ \dfrac{I}{4.0} \: = \: \dfrac{1}{u} \: + \: \dfrac{12}{ - 24} }}}$

Cross multiplying,

${\large: \longmapsto} \: \: {{\mathrm{ I \: = \: - \: \dfrac {12}{ 24} \: \times \: 4}}}$

We get,

${\large: \longmapsto} \: \: {{\mathrm{ I \: = \: - \: \dfrac {12}{ \cancel{24}} \: \times \: \cancel{4}}}}$

${\large: \longmapsto} \: \: \boxed{{\mathrm{I \: = \: \: - 2cm }}}$

$\underline{ \large{ \sf{Conclusion:-}}}$

$: \large{ \longmapsto} \: \mathrm{Position \: of \: image \: is 12 cm }$

$: \large{ \longmapsto} \: \mathrm{Size of image is -2 cm}$