Question

An object is kept at a far off distance in front of a convex lens of focal length 15 cm. State the position of the image.
9. A small object is placed in front of a convex lens of focal length 5 cm and a virtual image is formed at a distance of
25 cm. Find the magnification.
10. An object kept at a distance of 50 cm produces a virtual image of 10 cm in front of the lens. Draw a ray diagram to
show the formation of image and calculate the focal length. Is the lens is converging or diverging?
11. Where should an object be placed from a converging lens of focal length 10
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Answer 1

Answer:

9.) Then the magnification is 6.

Given data:

The lens given is a convex lens

Focal length f = 5 cm

The image formed is virtual so v = -25 cm.

Solution:

Using the lens formula, we can find the value of magnification

Substituting the values in the equation

u = 4.167 cm

To find magnification we have to use the relation

Where m is the value of magnification

Substituting the value

Therefore the image will be twice bigger as that of the object.

10.) First of all we find out the focal length of the lens. We know that the object is always placed in front of the lens on the left side, so the object distance is always taken as negative. Here the image is also formed in front of the lens on the left side, so the image distance will also be negative. Thus,

Object distance, u=−50cm (To the left of lens)

Image distance, v=−10cm (To the left of lens)

Focal length, f=? (To be calcualted)

Putting these values in the lens formula:

1v−1u=1f

We get: 1−10−1−50=1f

−5+150=1f

−450=1f

f=−504

So, Focal length, f=−12.5 cm

The minus sign for focal length shows that it is a concave lens. Please draw the ray diagram yourself.

We will not calculate the magnification produced by concave lens. We know that for a lens:

Magnification, m=vu

Here, Image distance, v=−10cm

And Object distance, u=−50cm

So, m=−10−50

m=+15

m=+0.2

Thus, the magnification produced by this concave lens is +0.2. Since the value of magnification is less than 1 (its is 0.2), therefore, the image is smaller than the object (or diminished). The plus sign for the magnification shows that the image is virtual and erect.