Question

An object of height 5 cm is placed at 10 cm in front
of convex lens of focal length 15 cm. Find the
height of the image, nature of image and position
of image.

Answer 1

Answer:

Given –

• Convex lens.
• Height of the object = 5 cm.
• Object Distance = -10 cm.
• Focal length = 15 cm.

To Find –

• Height of the Image.
• Nature of the Image.
• Position of the Image.

Solution –

Refer the attachment for the diagram.

Lens Formula : 1/v - 1/u = 1/f.

➸ 1/v - 1/-10 = 1/15

➸ 1/v + 1/10 = 1/15

➸ 1/v = 1/15 - 1/10

➸ 1/v = (10 - 15)/150

➸ 1/v = -5/150

➸ 1/v = -1/30

➸ v = -30 cm

Hence, the position of the image is 30 cm in front of the lens.

The nature of the image formed is virtual and erect.

The size of the image is enlarged.

Now, We know Magnification = v/u = h'/h.

Hence, -30/-10 = h'/5

➸ 3 = h'/5

➸ h' = 15 cm

Hence, height of the image formed is 15 cm.

Answer 2

Answer:

Explanation:

Given :-

Height of object = 5 cm.

Object Distance, u = -10 cm.

Focal length, f = 15 cm.

To Find :-

Height of the Image = ??

Nature of the Image = ??

Position of the Image = ??

Formula to be used :-

Lens Formula, 1/v - 1/u = 1/f

Magnification, m = v/u = h'/h

Solution :-

Putting all the values, we get

1/v - 1/u = 1/f.

⇒  1/v - 1/-10 = 1/15

⇒  1/v + 1/10 = 1/15

⇒  1/v = 1/15 - 1/10

⇒ 1/v = (10 - 15)/150

⇒  1/v = -5/150

⇒ 1/v = -1/30

⇒ v = -30 cm

Hence, the position of the image is 30 cm from of lens.

Now, Magnification = v/u = h'/h,

⇒ v/u = h'/h

⇒ -30/-10 = h'/5

⇒ h' = 15 cm

Hence, the height of the image is 15 cm.

And, the image is virtual and erect.