Question

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

Given:

Image distance (v) = +50 cm

Magnification (m) = -1  

[So the image is real and inverted]

We know that,

$\bf Magnification(m) = \frac{-v}{u}$

$\bf \implies u = \frac{v}{m} = \frac{50}{-1} = \boxed{-50 cm}$

So,

The needle is placed at 50 cm in front of the lens.

Now,

By the lens formula,

$\boxed{\frac{1}{f}=\frac{1}{v}-\frac{1}{u}}$

$\implies \frac{1}{f} = \frac{1}{50}-\frac{1}{(-50)}$

$\implies \frac{1}{f} = \frac{1}{50}+\frac{1}{50}$

$\implies \frac{1}{f} = \frac{1+1}{50}$

$\implies \frac{1}{f} = \frac{2}{50}$

$\implies \boxed{\frac{1}{f} =\frac{1}{25}}$

Therefore,

The focal length (f) = 25cm

Converting into m = 0.25 m

[As 1m = 100 cm]

Now,

$\boxed{\bf Power\ of\ the\ lens= \frac{1}{f(in\ m)}}$

So, the power of the lens is

$= \frac{1}{0.25} = 4D$

Hence,

The needle is placed at 50 cm in front of the lens.

And

The power of the lens is 4D.

Answer 2

$\mathfrak{\huge{\red{\underline{\underline{Answer :}}}}}$

The position of the image should be at 2F since the image is the real and same size.

It is given that the image of the needle is formed at a distance of 50 cm from the convex lens. Therefore, the needle is placed in front of the lens at a distance of 50 cm.

Object distance (u) = – 50 cm

Image distance, (v) = 50 cm

Focal length = ( f )

According to the lens formula,

$\frac{1}{v} - \frac{1}{u } = \frac{1}{f}$

$\frac{1}{f} = \frac{1}{50} - \frac{1}{ - 50}$

$= > \frac{1}{50} + \frac{1}{50} = \frac{1}{25}$

$f = 25cm \: = 0.25m \:$

$power \: of \: lens \: = \frac{1}{f(in \: metres \: )} = \frac{1}{0.25} = \: + 4d$

Hope it Helps !!