Question

an object 50 cm of length is placed at a distance of 25cm away from the converging lens of focal length 10cm draw the ray diagram

Answer 1

HEY BUDDY HERE IS UR ANSWER !!

Height of the object h = + 5 cm
Focal length f = + 10 cm
object-distance u = –25 cm
Image-distance v = ?
Height of the image h′ = ?



1 1 1
−
=
v u f
=> 1 1 1
=
−
v 10 25
=> 5 − 2 3
=
=
50 50
=> 50
v =
= 16.66 cm
3

A real and inverted image will be formed on other side of lense at 16.66 cm from its optical centre

=> v
m =
u
=> h2 16.66
=
h1 − 25
=> h2 16.66
=
5 − 25
=> −16.66 × 5
h2 =
25
=>
h2 = − 3.33 cm
An inverted, 3.33 cm high image will be formed.

Hope u got my process !!

》》BE BRAINLY 《《

Answer 2

Given, height of object = 5cm

Position of object, u = - 25cm

Focal length of the lens, f = 10 cm

Hence, position of image, v =?

We know that,

1/v - 1/u = 1/f

1/v + 1/25 = 1/10

So, 1/v = 1/10 - 1/25

S0, 1/v = (5 - 2)/50

That is, 1/v = 3/50

So, v 50/3 = 16.66 cm

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

Now, magnification = v/u

That is, m = 16.66/-25 = -0.66

Also, m= height of image/height of object

Or, -0.66 = height of image / 5 cm

Therefore, height of image = -3.3 cm

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

Thus, position of image = At 16.66 cm on opposite side of lens

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