Question

An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of length 10 cm. If the distance of the object from the optical centre of the lens is 20 cm, determine the position, nature and size of the image formed using the lens formula.​

Answer 1

Explanation:

In the question we have

f = -10cm

u = -20cm

object height = 5cm

We need to find out

v = ???

Using lens formula

1/f = 1/v - 1/u

It can also be written as

=> 1/v = 1/f + 1/u

=> 1/v = -1/10 + (-1/20)

=> 1/v = -2-1/20

=> 1/v = -3/20

=> v = -20/3

=> v = -6.33cm

Now again using formula

m = hi/ho = v/u

Formula can be also written as

hi = v/u x ho

=> 20 x 5/3 x 20

=> 5/3

=> -1.6cm

Some results-

• Concave lens Also forms virtual images.
• It will be diminished.

Answer 2

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Given:

• We have been given that the height of an object is 5cm.
• The object is placed perpendicular to the principal axis of a concave lens of length 10cm.
• The distance of the object from optical center is 20cm.

To Find:

• We need to find the position, nature and size of the image formed using the lens formula.

Solution:

We have been given that h = 5cm,

f = (-10cm), and u = (-20cm).

We can calculate the image distance(v) by the lens formula. We have,

1/f = 1/v - 1/u

=> 1/v = 1/f + 1/u

Substituting the values, we have

1/(-10) + 1/-(20)

Therefore, 1/v = -1/10 - 1/20

= (-20 - 10)/200

= -30/200

= -3/20

Therefore, v = -20/3cm.

The image is situated at a distance of -20/3cm. It is situated on the same side of lens. Hence, the image formed is virtual.

Now, we need to find the magnification of the lens, we have

m = h'/h = v/u

=> h' = (v/u) × h

= [-20/{3 × (-20)}] × 5

= 5/3

= 1.67cm

Hence, the image formed is virtual and erect.