Question

if an object of 7cm height is placed at a distance of 12 cm form a convex lens of focal length 12cm find the position ,nature and height of the image?

Answer 1

S O L U T I O N :

Given :

• Height of object, (h1) = 7 cm
• Distance of object, (u) = -12 cm
• Focal length, (f) = 12 cm

Explanation :

As we know that formula of the lens :

• 1/f = 1/v - 1/u

According to the question :

➝ 1/f = 1/v - 1/u

➝ 1/12 = 1/v - (-1/12)

➝ 1/12 = 1/v + 1/12

➝ 1/12 = 1/v + 1/12

➝ 1/v = 1/12 - 1/12

➝ 1/v = 1-1/12

➝ 1/v = 0/12

➝ v × 0 = 12

➝ v = 12 cm

Therefore, the height of image will be 12 cm .

Now,

As we know that Formula of magnification:

• Magnification = Height of Image/Height of object = -v/u

➝ h2/h1 = -v/u

➝ h2/7 = -12/-12

➝ h2/7= 1

➝ h2 = 7 cm

Thus,

The image is virtual & erect and the size of image is 7 cm .

Answer 2

Given :-

Focal length = 12 cm

Distance = -12 cm

Height = 7 cm

To Find :-

Position and nature and height

Solution :-

As we know that

$\huge \tt \: \frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$\sf \: \dfrac{1}{12} = \dfrac{1}{v} + \dfrac{1}{-12}$

$\sf \: \dfrac{1}{12} = \dfrac{1}{v} + \dfrac{1}{12}$

$\sf \: \dfrac{1}{v} = \dfrac{1}{12} - \dfrac{1}{12}$

$\sf \: \dfrac{1}{v} = \dfrac{1 - 1}{12} = \dfrac{0}{12}$

$\sf \: v = 12$

Hence :-

The height of image is 12 cm

Now,

Let's find magnification

$\sf \: \dfrac{ho}{hi} = \dfrac{ - v}{u}$

$\sf \: \dfrac{7}{hi} = \dfrac{ - 12}{-12}$

$\sf \: \dfrac{7}{hi} = 1$

$\sf \: hi = 7 \: cm$

Hence :-

Height of image is 7 cm and it is virtual and erect.