An object of height 4.0 cm is placed at a distance of 30 cm from optical centre 'O' of a convex lens of focal length 20 cm. Draw a ray diagram to find the position and size of the image formed. Mark optical centre 'O' and principal focus 'F' on the diagram. Also find the approximate ratio of size of image to the size of object.
An object of height 4.0 cm is placed at a distance of 30 cm from optical centre 'O' of a convex lens of focal length 20 cm. Draw a ray diagram to find the position and size of the image formed. Mark optical centre 'O' and principal focus 'F' on the diagram. Also find the approximate ratio of size of image to the size of object.
Let the focal length be f. Let u and v be the distance of the object and image from focus respectively. Let ho and hi be the heights of the object and the image respectively.
Given, f =+ 20 cm, u= - 30 cm, ho = 4 cm
We know that,
1f=1v−1u
120=1v−1−30
1v=160
⇒v=60 cm
So, hiho = vu
hi4=60−30
⇒hi=−8 cm
Thus, the height or size of the image is 8 cm. The minus sign shows that this height is in the downward direction, that is, the image is formed below the axis.
Ratio of size of image to object = 8 : 4 = 2 : 1
So image is enlarged and it is formed beyond 2F2.
The object is placed between F1 and 2F1
Image is formed beyond 2F2 and it is real and inverted.
Let the focal length be f. Let u and v be the distance of the object and image from focus respectively. Let ho and hi be the heights of the object and the image respectively.
Given, f =+ 20 cm, u= - 30 cm, ho = 4 cm
We know that,
1f=1v−1u
120=1v−1−30
1v=160
⇒v=60 cm
So, hiho = vu
hi4=60−30
⇒hi=−8 cm
Thus, the height or size of the image is 8 cm. The minus sign shows that this height is in the downward direction, that is, the image is formed below the axis.
Ratio of size of image to object = 8 : 4 = 2 : 1
So image is enlarged and it is formed beyond 2F2.
The object is placed between F1 and 2F1
Image is formed beyond 2F2 and it is real and inverted.
