Question

An Object Of Height 4 Cm Is Placed On The Principal Axis At A Distance Of 20 Cm In Front Of A Convex Lens Of Focal Length 30 Cm. Find The Distance Of The Image From The Lens,Its Height And Nature.
Please Answer Fast..

Answer 1

Answer:

ANSWER

The relationship between the radius of curvature and focal length is that the radius of curvature is twice the focal length.

In this case, focal length of the concave mirror is given as 15 cm. Hence, the radius of curvature is 15×2=30 cm.

When the object is placed on the centre of curvature, on the principal axis, in front of a concave lens, the image is formed at the same point. The image formed is real and inverted.

So, in this case, the object distance is equal to the image distance.

Therefore, the object size is equal to the image size.

Answer 2

The distance of the image from the lens is -60 cm

12cm is the object's height and the nature of the image is upright and virtual

Step-by-step process

Given:

object height, Ho= 4cm

object distance ;u= -20cm

Focal length;f= 30cm

Calculations:

Using the lens formula, we can find the image distance(v):

1/v - 1/u = 1/f

Putting the values in the equations

1/v- 1/-20 = 1/30

1/v = 1/30+1/-20

1/v =2-3/60

v = -60cm

So, the image is formed 60cm to the left of the lens.

Using magnification formula

m = v/u

m = -60/-20

m = 3

Thus, the image will be thrice more magnified than the object

The height of the object is given by

m = hi/ho

3= Hi/4

Hi= 12 cm

The height of the image is 12 cm.

Result:

The distance of the image from the lens is -60 cm

The height is 12cm and the nature of the image is upright and virtual.

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