Physics, asked by kumawatmayank133, 1 year ago

the image of an object formed by a mirror is real,inverted and is of magnification -1.If the image is at the distance of 30cm from the mirror, where is the object placed? Find the position of the image if the object is now moved 20cm towards the mirror. what is the nature of the image obtained? Justify your answer with the help of ray diagram.

Answers

Answered by AbhijithPrakash
23

Given the image formed is real inverted and its magnification is -1

If magnificaion is -1 then it means the nature of the image is real invertedand of the same size . in this case the object is placed at C and image is formed at C.

If the image is at a distance of 30cm from the mirror then the object is also at a distance of 30cm from the mirror.

If the object is moved 20cm towards the mirror then the position of the object is between the pole and the principal focus  (f=15cm)

When the object is moved 20cm towards the mirror then the position of the image will be by the mirror formula, and the nature of the image is .


look the attachment for the diagram

Attachments:

ktejpalsingh: nice one it is helpful
AbhijithPrakash: Thanks bro...
Mohddaudkhan: the ray diagram is wrong
Mohddaudkhan: because it form virtual and erect image not real and inverted
Mohddaudkhan: question given that the image is real and inverted but u draw a ray diagram of virtual and erect image
Mohddaudkhan: srry u are right i was wrong
Answered by BendingReality
9

Answer:

+ 30 cm

Explanation:

Given :

Magnification = - 1

Image distance = 30 cm

We know

m = - v / u

- 1 = 30 / u

u = - 30 cm

Now we have ,

1 / f = 1 / v + 1 / u

1 /  f = 1 / - 30 - 1 / 30

f = - 15 cm

When object moved 20 cm

u' = - 30 + 20 = - 10 cm

We have f = - 15 cm

We have to find new image distance v'

Again ,

1 / f = 1 / v' + 1 / u'

- 1 / 15 = 1 / v' - 1 / 10

1 / v' = 1 / 30

v' = 30.

Nature of image as :

Virtual

Erect

Magnified.

Attachments:
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