Question

A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also find the power of the lens.

Answer 1

Solution :-

A convex lens forms a real and inverted image of a needle at a distance of 50cm from it.

Therefore,

Magnification = -1

Image distance ( v) = 50 cm

Now,

As we know that,

m = h1/h2 = v/u

Put the required values,

-1 = 50 / u

u = -50 cm

Thus, The object distance is -50 cm

Now ,

The image is equal to the size of the object

It means object is placed at twice the focal length

Therefore,

Focal length = 50/2

Focal length = 25cm

Focal length = 0.25m

[ SI unit of distance is m ]

Now,

As we know that,

Power of lens = 1 / Focal length

P = 1/0.25

P = 100/25

P = 4D

Hence, The power of the lens is

4 diptore $.$

Answer 2

Given:

• The image formed by convex lens is real and inverted.
• Distance of an image is 50cm.

To Find:

• Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also find the power of the lens.

Solution:

We know that,

The image formed by convex lens is real and inverted.

• So, Magnification = -1.

Image distance, v = 50cm.

Here,

$: \implies \sf m = \dfrac{h_1}{h_2} = \dfrac{v}{u}$

$: \implies \sf -1 = \dfrac{50}{u}$

$: \implies \sf u = -50cm.$

$\large{\blue{\underline{\boxed{\bf u = -50cm}}}}$

Now,

Image is equal to size of object.

• So, Focal length = $\sf \dfrac{50}{2}$
• Focal length = 25cm.
• Converting cm into m. [Divide by 100]
• Focal length = 0.25m.

We know,

$: \implies \sf Power \: of \: lens, P \: = \: \dfrac{1}{focal \: length}$

$: \implies \sf P = \dfrac{1}{0.25}$

$: \implies \sf P = \dfrac{100}{25}$

$\large{\red{\underline{\boxed{\bf P = 4D}}}}$

Hence:

• Needle should be placed -50cm in front of the convex lens if the image is equal to size of the object.
• The power of lens is 4 D.