Question

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed. [ CBSE 2011 ]

Class :- X

Light : Reflection and Refraction chapter

Answer 1

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



Nature of image – Real and inverted....

Answer 2

Hello Dear Soul ❤

Your Answer :

Height of the object, h' = 5 cm

Distance of object from conversing lens, u = 25 cm

Focal length of conversing lens, f = 10 cm

Using lens formula,

1/v - 1/u = 1/f

1/v = 1/f + 1/u

1/v = 1/10 - 1/25 = 15/250

v = 250/15 = 16.66 cm

Also for conversing lens,

h''/h' = v/u { h'' = image height }

h'' = (v/u)× h' = (16.66 × 5) / (-25) = -10/3 = -3.3 cm

Thus, The image is inverted and formed at a distance of 16.7 cm behind the lens and measured 3.3 cm.

Note : Ray diagram of this problem is given in attached file.

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