Question

A object of 3 cm h is placed on the principle axis and perpendicular to it a real image at 60 cm is formed by convex lens the radius of curvature of lens is 40 cm then calculate 1 power of lens 2 position of object 3 height of image?....

Answer 1

Given information :

Height of object ( O ) = 3 cm

Height of image ( I ) = ?

Image distance ( v ) = 60 cm

Radius of curvature ( r ) = 40 cm

Object distance ( u ) = ?

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Finding the power of lens

Power = 1 / f

Focal length ( f ) = 2 × r

                           = 2 × 40 cm

                           = 80 cm

                           = 80 / 100  m

Power = 1 / f

           = 1 / 80 / 100

           = 100 / 80

           = 10 / 8

           = 5 / 4

           = 1.25

The power is + 1.25 D

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Finding the position of object

Now , the image is real . So it is a convex lens.

Lens formula : 1/f = 1/v - 1/u

Now : for convex lens :

• u is negative
• v is positive ( real image )
• f is positive

So :

1/f = 1/v - ( - 1/u )

==> 1/f = 1/v + 1/u

==> 1/u = 1/f - 1/v

==> 1/u = 1/80 - 1/60

==> 1/u = ( 3 - 4  ) / 240

==> 1/u = - 1/240

==> u = -240 [ u is -ve for convex lens ]

Hence

The position of the object is 240 cm behind the lens

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Finding the height

I / O = v / u

==> I / 3 = 60 / 240

==> I / 3  = 1 / 4

==> I = 3 / 4

==> I = 0.75 cm

The height of image is 0.75 cm

Hope it helps ^__^

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Answer 2

In the questions given

Height of objects 3 cm . and image distance 60 cm.Radius of curvature 40 cm.

In the questions we find .

power of lens = ?

Height of images = ?

Objects distance = ?

Now we find focal length of lens =

Focal length = 2 radius of curvature

F = 2r

F =2× 40

F = 80 cm

Here we find Power .
$Power = \frac{1}{f(in \: m)}$
Now we change in metres.

$80cm = \frac{80}{100m} \\ \\ power = \frac{1}{ \frac{80}{100} } \\ \\ = power = \frac{100}{80} \\ \\ \frac{10}{8} \\ \\ = 1.25Diopter$

From the property of convex lens we using lens formula.


$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \\ putting \: the \: value \: in \\ \: formula \\ \\ \frac{1}{u} = \frac{1}{80} - \frac{1}{60} \\ \\ \frac{1}{u} = \frac{3 - 4}{240} \\ \\ \frac{1}{u} = \frac{ - 1}{240} \\ \\ u \: = - 240 \\ \\ \\ \\ \frac{Height \: of \: images }{ Height of objects } = \frac{v}{u} \\ \\ \frac{i}{3} = \frac{60}{240} \\ \\ \frac{i}{3} = \frac{1}{4} \\ \\ 4i = 3 \\ \\ i = \frac{3}{4} \\ \\ i = 0.75cm.$


The height of the image is 0.75cm.


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