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.
​

Answer 1

ANSWER

Given,

Height of object =5cm

Position of object, u=−25cm

Focal length of the lens, f=10cm

Position of image, v=?

We know that,

1/v - 1/u = 1/f

1/v + 1/25 = 1/10

1/v = 1/10 - 1/25

So ,

1/v = (5-2)/50

That is ,

1/v =3/50

So ,

v = 50/3 = 16.66 cm

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

Now, magnification = v/u

That is,

m = 16.66/-25 = −0.66

Also,

m =  heoght of image / height of object

OR

-0.66 = height of image/5 cm

Therefore, Height of image = 3.3 cm

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.

​

Answer 2

Explanation:

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