Question

a canvex lens forms a real and inverted image of a needle at a distance of 50 from it where is the needle placed in front of the canvex lens if the image is equal to the size of the object? also find the power of the lens​

Answer 1

Explanation:

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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.

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