Question

3. 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

$\huge\underline\mathfrak\red{Question}$

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.

$\huge\underline\mathfrak\orange{Answer}$

Object distance, u= – 50 cm

Image distance, v= 50 cm

Focal length = f

According to the lens formula,

u = -50 cm v = + 50 cm.

f =?

1/f = 1/v €“ 1/u 1/f

= 1/50 + 1/50 f = + 25 cm.

= 0.25 m Power of lens (P)

= 1/f = 1/ 0.25

= + 4D.

$\huge\underline\mathcal\purple{BTS}$

Answer 2

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.

• The power of the lens is 4D.