Question

An object of height 10 cm is placed at a distance of 25 cm from a convex lens of focal length 15 cm . Find position nature and size of the image
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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

Height of the object. (h )=10cm

Distance of object from a convex lens( u. )= -25cm

Focal length=15cm

From the relation,

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

$= > \frac{1}{v} - \frac{1}{ - 25} = \frac{1}{15}$

$= > \frac{1}{v} = \frac{1}{15} - \frac{1}{25}$

Taking LCM of 15 and 25 we get,

$= > \frac{1}{v} = \frac{5 - 3}{75}$

$= > \frac{1}{v } = \frac{2}{75}$

$= > 2v = 75 \\ = > v = \frac{75}{2} \\ = > 37 \frac{1}{2}$

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