Question

An object 3cm high is placed 24cm away from converging lence if focal length 8cm find position, nature,and height of image

Answer 1

Explanation:

(i) h1 = 3 cm

u = -24 cm

f = 8 cm

Lens formula : 1/v – 1/u = 1/f

1/v – 1/-24 = 1/8

1/v = 1/12

V = 12 cm

Image is formed 12 cm behind the lens.

m = v/u = h2/h1

12/-24 = h2/3

h2 = -1.5 cm

Image is 1.5 cm high, real and inverted.

(ii) u = - 3 cm

h1 = 3 cm

f = 8 cm

Lens formula : 1/v – 1/u = 1/f

1/v – 1/-3 = 1/8

1/v = - (5/24)

V = -4.8 cm

Image is formed 4.8 cm in front of the lens.

m = v/u = h2/h1

-4.8/-3 = h2/3

h2 = + 4.8 cm

Image is 4.8 cm high, virtual and erect.

hope u will be satisfied with my answer...

Answer 2

GiveN :

• Object Height (Ho) = 3 cm
• Object Position (u) = - 24 cm
• Focal length (f) = 8 cm

To FinD :

• Nature, size and position of lens

SolutioN :

Use Lens formula :

➠ 1/f = 1/v - 1/u

⇒1/v = 1/f + 1/u

⇒1/v = 1/8 - 1/24

⇒1/v = 3 - 1/24

⇒1/v = 2/24

⇒1/v = 1/12

⇒v = 12

✯ Image is formed at 12 cm away from lens

$\rule{200}{2}$

Now, use formula for magnification :

➠ m = v/u

⇒m = 12/-24

⇒m = -1/2

⇒ m = - 0.5

And also we know that :

⇒m = h'/ho

⇒-0.5 = h'/3

⇒h' = -0.5 × 3

⇒ h' = -1.5

So, image size is 1.5 cm. Negative sign shows image is in below the principal axis.

• Nature of Image is Real and Inverted .