Please Answer this....
Class 10 science board cbse 2020
Qn. 29

a) See attachment

b) -180 cm, virtual image

c) Magnification: 10, image is 10 times the size of the object, upright

Explanation:

a)

The ray diagram for this situation is shown in attachment.

In order to find the position of the image, we proceed as follows:

- We draw a ray of light going from the tip of the object towards the lens, parallel to the principal axis - this ray is refracted towards the principal focus on the other side

- We draw another ray of light going from the tip of the object towards the centre of the lens

- We prolong both rays: we see that they don't meet on the right side of the lens. Therefore, we prolong them on the left side, where they meet - this means that the image is virtual, because it cannot be projected on a real screen (it is formed on the same side as the object).

b)

To find the nature and the position of the image, we use the lens equation:

where:

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

is the focal length

is the object distance

Solving for q, we find the position of the image:

The negative sign indicates that the image is virtual (on the same side of the object).

c)

The magnification is given by:

where

q is the image distance from the lens

p is the object distance from the lens

Here we have

So the magnification is:

This means that the image size is a factor 10 times the object size. In fact, we can write

where

y' is the image size

y is the object size

Substituting,

So, the image is 10 times the object (in size), and it has the same orientation, upright (because of the positive sign).