Question

find the power of convex lens which forms a real and inverted image of magnification of -1of an object placed at a distance of 20 cm from its optical centre​

Answer 1

Answer:

20cm object placed at distance

Answer 2

Answer:

Given:

Image distance (v) = +50 cm

Magnification (m) = -1

[So the image is real and inverted]

We know that,

$\bf Magnification(m) = \frac{-v}{u}</p><p>$

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

So,

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

Now,

By the lens formula,

$\boxed{\frac{1}{f}=\frac{1}{v}-\frac{1}{u}} </p><p>$

$\implies \frac{1}{f} = \frac{1}{50}-\frac{1}{(-50)}</p><p>$

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

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

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

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

Therefore,

The focal length (f) = 25cm

Converting into m = 0.25 m

[As 1m = 100 cm]

Now,

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

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.