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When an object is at a distance of twice the focal length from a concave lens, the image produced is virtual and smaller than the object. What happens to the image if the object is shifted closer to the lens to a point one focal length away from it?

The image produced is virtual and enlarged.
The image produced is virtual and smaller than the object.
The image produced is real and enlarged.
The image produced is real and smaller than the object.
The image produced is virtual and of the same size as the object.

Respuesta :

Hagrid
The right answer for the question that is being asked and shown above is that: "The image produced is virtual and of the same size as the object." the image if the object is shifted closer to the lens to a point one focal length away from it is that The image produced is virtual and of the same size as the object.

Answer:

The image produced is virtual and smaller than the object.

Explanation:

For concave lens we know that

[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}[/tex]

here we have

[tex]\frac{1}{d_i} + \frac{1}{2f} = -\frac{1}{f}[/tex]

[tex]\frac{1}{d_i} = - \frac{3}{2f}[/tex]

[tex]d_i = -\frac{2f}{3}[/tex]

Magnification will be given as

[tex]M = \frac{d_i}{d_o}[/tex]

[tex]M = -2/3[/tex]

so image will be virtual and formed behind the lens

Now the object position is shifted to new position at distance of focal length

now again we will have

here we have

[tex]\frac{1}{d_i} + \frac{1}{f} = -\frac{1}{f}[/tex]

[tex]\frac{1}{d_i} = - \frac{2}{f}[/tex]

[tex]d_i = -\frac{f}{2}[/tex]

Magnification will be given as

[tex]M = \frac{d_i}{d_o}[/tex]

[tex]M = -1/2[/tex]

So again we will have virtual image with magnification 1/2

so here size of image is less than object size by factor of 1/2 and it is virtual

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