Question

A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also find its magnification. ​

Answer 1

Answer:

Object's size (h1) = 2 cm

Focal length of convex lens (f) = 10 cm

Object distance from the lens (u) = -15 cm

Image distance ( v) = ?

Image size (h2) = ?

We know,

1/v - 1/u= 1/f

1/v = 1/f + 1/u

1/v = 1/10 - 1/15

1/v = 1/30

Thus, v = 30 cm

Now, v = 30 cm, 'v' is positive, that means the image is formed on the right side of the lens. That's why it is REAL and INVERTED.

Now, magnification =?

Linear magnification (m) = Size of image / Size of object = h2 / h1 = v/u

m = h2/2 = 30/-15

m= h2 × -15 = 30×2

m = -15 h2 = 60

h2 = 60 / -15

h2 = -4 cm

Size of the image = -4 cm

It's in negative , that means that the image is INVERTED.

Explanation:

Answer 2

★ QUESTION :

A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also find its magnification.

GIVEN :

• Height of the object (h) = + 2.0 cm
• Focal length (f) = + 10 cm
• Distance of the object (u) = - 15 cm

TO FIND :

• Distance of the image (v) = ?
• Height of the image (h') = ?

STEP - BY - STEP EXPLAINATION :

➠ Height of the object (h) = + 2.0 cm (GIVEN)

➠ Focal length (f) = + 10 cm (GIVEN)

➠ Distance of the object (u) = - 15 cm (GIVEN)

Now, here we will have to find the distance of the image (v) and height of the image. (h')

Therefore, as we know that,

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

$⟹ \: \frac{1}{v} = \frac{1}{( - 15)} + \frac{1}{10} = - \frac{1}{15} + \frac{1}{10}$

$⟹ \: \frac{1}{v} = \frac{ - 2 + 3}{ \: 30} = \frac{1}{30}$

So, here the value of Distance of the image (v) = +30 cm

The images formed at the distance of 30cm on the other side of the optical centre. The nature of the image is Inverted and Real.

⟹ ᴍᴀɢɴɪғɪᴄᴀᴛɪᴏɴ (m) = h' / h = v / u

⟹ Height of the image, h' = (2.0) + (+30 / -15) = - 4.0 cm

⟹ m = +30 cm / -15 cm = -2

So, the nature of image is inverted and real. It is formed below the principal axis. It consists of 4 cm tall, and it is formed at a distance of 30 cm on on the other side of the lens.