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).

the force on the thing and the distance the force moves the thing

the answer for apex is the lowest