question_answer

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.

Answer:

Given, \[f=+20cm,\,u=-30\,cm,\,{{h}_{o}}=4\,cm\]

We know that,

\[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\]

\[\frac{1}{20}=\frac{1}{v}-\frac{1}{-30}\]

\[\frac{1}{v}=\frac{1}{60}\]

\[\Rightarrow \]   \[v=60cm\]

So,       \[\frac{{{h}_{i}}}{{{h}_{o}}}=\frac{v}{u}\]

\[\frac{{{h}_{i}}}{4}=\frac{60}{-30}\]

r            \[{{h}_{i}}=-8cm\]

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 \[=-2\]

So image is enlarged beyond\[2{{F}_{2}}\].

Object between \[{{F}_{1}}\] and \[2{{F}_{1}}\]

Image is formed beyond \[2{{F}_{2}},\] real, inverted.