Question

$\mathtt{ \underbrace {\Huge{\color{red}\bigstar{ \: Question}}}}$
An object 5 centimetre in length is held 25cm away from a converging lens of focal length 10cm .Draw the ray diagram and find the position size and the nature of image formed.​

Answer 1

Answer:

Height of object =5cm

Position of object, u=−25cm

Focal length of the lens, f=10cm

Position of image, v=?

We know that,

v

1

−

u

1

=

f

1

v

1

+

25

1

=

10

1

v

1

=

10

1

−

25

1

So,

v

1

=

50

(5−2)

That is,

v

1

=

50

3

So,

v=

3

50

=16.66cm

Thus, distance of image

is 16.66cm on the opposite side of lens.

Now, magnification =

u

v

That is,m=

−25

16.66

=−0.66

Also,

m=

heightofobject

heightofimage

or

−0.66=

5cm

heightofimage

Therefore, Height of image =3.3cm

The negative sign of the height of image shows that an inverted image is formed.

Thus, position of image is at 16.66cm on opposite side of lens.

Size of image =−3.3cm at the opposite side of lens

Nature of image is real and inverted.

solution

Explanation:

Hope this will help you,Please Follow Me

Answer 2

$\mathtt{ \underbrace {\Huge{\color{red}\bigstar{Solution}}}}$

Given,

Given,Height of object =5cm

Given,Height of object =5cmPosition of object, u=−25cm

Given,Height of object =5cmPosition of object, u=−25cmFocal length of the lens, f=10cm

Given,Height of object =5cmPosition of object, u=−25cmFocal length of the lens, f=10cmPosition of image, v=?

Given,Height of object =5cmPosition of object, u=−25cmFocal length of the lens, f=10cmPosition of image, v=?We know that,

v1 − u1 = f1

v1 − u1 = f1

+ 251

= 101v1

= 101v1 = 101

− 251

So,

v1

v1 = 50(5−2)

That is,

v1

= 503

So,

v= 350

v= 350 =16.66cm

=16.66cmThus, distance of image is 16.66cm on the opposite side of lens.

=16.66cmThus, distance of image is 16.66cm on the opposite side of lens.Now, magnification = uv

=16.66cmThus, distance of image is 16.66cm on the opposite side of lens.Now, magnification = uv

=16.66cmThus, distance of image is 16.66cm on the opposite side of lens.Now, magnification = uv

That is,

m= −2516.66

=−0.66

=−0.66Also,

=−0.66Also,m= heightofobject

=−0.66Also,m= heightofobjectheightofimage

=−0.66Also,m= heightofobjectheightofimage or

−0.66= 5cm

−0.66= 5cmheightofimage

−0.66= 5cmheightofimageTherefore, Height of image =3.3cm

−0.66= 5cmheightofimageTherefore, Height of image =3.3cmThe negative sign of the height of image shows that an inverted image is formed.

−0.66= 5cmheightofimageTherefore, Height of image =3.3cmThe negative sign of the height of image shows that an inverted image is formed.Thus, position of image is at 16.66cm on opposite side of lens.

−0.66= 5cmheightofimageTherefore, Height of image =3.3cmThe negative sign of the height of image shows that an inverted image is formed.Thus, position of image is at 16.66cm on opposite side of lens.Size of image =−3.3cm at the opposite side of lens

−0.66= 5cmheightofimageTherefore, Height of image =3.3cmThe negative sign of the height of image shows that an inverted image is formed.Thus, position of image is at 16.66cm on opposite side of lens.Size of image =−3.3cm at the opposite side of lensNature of image is real and inverted.