Question

An object is placed at a distance of 0.5 m from a concave lens of a focal length 0.2 m Find the nature and position of image ?

Answer 1

Answers :-

Given

u = –0.5 m
= –50cm

f = –0.2m (concave lens)

= –20cm


According to lens formula ,

1/v –1/u = 1/f

1/v = 1/u +1/f

1/v = –1/50 –1/20

= –2–5/100

= –7 /100

Then , V = –100/7

V = –14.3

Thus , the negative sign of indicates that the image is formed at a distance of 14.3 cm on the left side of the lens .

The image is formed is virtual and erect

Magnification = v/u

= 100/ 7 * (–50)

= +2/7
= 0.3

The image formed by the diverging lens is diminished

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Answer 2

Solution :

Object will be beyond Centre of curvature.
So,
Image will be between focus and centre of curvature.
It will be real, inverted and diminished (small sized) image.
C = 2f

f is the focus.

Now , u = -0.5 m
f = -0.2 m

-» u sin g the lens formula
1/-0.5 = 1/v - 1/-0.2

1/-0.5 × 0.1x = 1/x × 0.1x - 1/-0.2 × 0.1x

= 0.1/-0.5x = 0.1 + 0.1/0.2x

= -0.1/0.5x - 0.1/0.2x = 0.1 + 0.1/0.2x - 0.1/0.2x

= -0.2x - 0.1/0.2x = 0.1

= -0.2x - 0.5x = 0.1

= - 0.7x = 0.1

= -0.7x/-0.7 = 0.1/-0.7

= x = 0.1/0.7

x = 1/7

And then = 1/7/-0.5

-1/3.5 is the final answer.

Or -2/7

The image formed is diminished, Virtual and erect .

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