Question

A convex lens forms a real and inverted image of a needle at a distance of 50 cm from the lens if the image of the same size as the new year is the needle placed in front of the lens also find the power of the lens

Answer 1

Given,

focal length, f = -15 cm

Distance of image, v = -10 cm

Distance of object, u =?

We know that,

1/v - 1/u = 1/f

So, 

1/-10 - 1/u = 1/-15

So, -1/10 + 1/15 = 1/u

(-3 + 2)/30 = 1/u

-1/30 = 1/u

Therefore u = -30 cm

Thus object is placed 30 cm away from the concave lens.

 Negative sign shows that object is at 30cm in front of the lens.

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 !!