Question

A pin of length 2.00 cm is placed perpendicular to the principal axis of a converging lens. An inverted image of size 1.00 cm is formed at a distance of 40.0 cm from the pin. Find the focal length of the lens and its distance from the pin.

Answer 1

The focal length of the lens f = 8.9 cm and its distance from the pin is u = 26.66 cm.

Explanation:

Given data:

Distance between object and image = (-u) + v = 40 cm

Size of pin ho =2 cm

h1= 1 cm

Since hi/ho = v/-u = Magnification

−2/2 = −v/u

u = 2v

Also gives.

3 v=40 cm

v = 13.3=40/3 cm,

u = 80/31/v-1/u = 1/f

3/40 - 3/-80 = 1/f

1/f = 3/40 [1  + 1/2] = 3/40 x 3/2

f = 80/9 = 8.9 cm

f = 8.9 cm

u = 26.66 cm  

The focal length of the lens f = 8.9 cm and its distance from the pin is u = 26.66 cm.

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Answer 2

The focal length of the lens f = 8.89 cm and its distance from the pin is 26.66 cm.

Explanation:

Given data in the question  :

H =  2 cm        

h' = 1 cm  

H is pin of length

h’ is image of size

Inverted image = Real image

$m=-\frac{v}{u}=\frac{h^{\prime}}{h}$    

   $=-\frac{1}{2}$

u = - 2 v      

Using lens formula :

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

$\frac{1}{v}-\frac{1}{-2 v}=\frac{1}{f}$

$\frac{1}{v}+\frac{1}{2 v}=\frac{1}{f}$

$f=\frac{2 v}{3}$   -----> eqn (1)

Where, U is positive, and v is negative

-u + v = 40 cm

3 v = 40  cm    

$v=\frac{40}{3} \mathrm{cm}$

$u=-\frac{80}{3} \mathrm{cm}$              

    $=\frac{3}{40}+\frac{3}{80}=\frac{9}{80}$    

$f=\frac{80}{9} \mathrm{cm}$

Focal length f = 8.89 cm (or) 8.9

From eqn (1),

$f=\frac{2 v}{3}$  

v = $\frac{f \times 3}{2}$

v = $\frac{8.9 \times 3}{2}$

v = 13.3    

2 v = 26.66 cm

Thus, the focal length is f = 8.9 cm and its distance is 26.66 cm.