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A concave lens has focal length f. A real object placed at a distance f in front of the lens from the pole produces an image a) at infinity b) at f c) at f/2 d) at 2/f

A concave lens has focal length f. A real object placed at a distance f in front of the lens from the pole produces an image a) at infinity b) at f c) at f/2 d) at 2/f

Grade:upto college level

3 Answers

Saurabh Koranglekar
askIITians Faculty 10335 Points
3 years ago
The image will be a virtual image (behind the mirror) between the pole and the focus.

The position of the image can be calculated from the well-known mirror formula:

1/f = 1/do + 1/di where “do” is the distance of the object from the pole and “di” is the distance of the image from the pole.

You need to be careful with the signs (+ or -) for the distances. Unfortunately there is more than one “sign convention”. What I will use here is the convention that virtual image distances are negative and the focal length of a convex mirror is also negative.

So, the focal length isminusf and the object distance isplusf.

Hence: -1/f = 1/f + 1/di.

Adding -1/f to both sides of the equation gives -2/f = 1/di. Therefore di = -f/2.

So the image is at a distance f/2 behind the mirror
Vikas TU
14149 Points
3 years ago
Focal length of convex mirror as f
           Distance of the object from the mirror = u
           Distance of the image from the mirror = v
The convex mirror with object placed in front of it at a distance from the pole forms an image by it whose distance  is given by the formula.
As we know that the sign convention is used, so we must know whether the image is real or virtual.
Then we may apply the optic formula as given 1/v + 1/u = 1/f 
1/v = 1/f - 1/u 
1/f +1/f = 2/f 
Arun
25750 Points
3 years ago
Object distance  u=−2f​
Focal length of convex mirror is  f.
Using mirror image  :    v1​+u1​=f1​
Or    v1​+−(f/2)1​=f1​
⟹ v=3f​
Magnification     m=u−v​=−f/2−(f/3)​=32​
The image formed is virtual and erect and diminished.

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