Lakhmir Singh and Manjit Kaur Solutions for Chapter: Refraction of Light, Exercise 28: Long Answer Type Questions

Find the position and size of the virtual image formed when an object $2 \mathrm{cm}$ tall is placed $20 \mathrm{cm}$ from: A diverging lens of focal length $40 \mathrm{cm}$.

Find the position and size of the virtual image formed when an object $2 \mathrm{cm}$ tall is placed $20 \mathrm{cm}$ from: A converging lens of focal length $40 \mathrm{cm}$.

Draw labelled ray diagrams to show the formation of images when: (i) an object $2 \mathrm{cm}$ tall is placed $20 \mathrm{cm}$ from a diverging lens of focal length $40 \mathrm{cm}$ and (ii) an object $2 \mathrm{cm}$ tall is placed $20 \mathrm{cm}$ from a converging lens of focal length $40 \mathrm{cm}$. (The diagrams may not be a according to scale).

A small object is placed $150 \mathrm{mm}$ away from a diverging lens of focal length $100 \mathrm{mm}$.

Copy the figure below and draw rays to show how an image is formed by the lens.

A small object is placed $150 \mathrm{mm}$ away from a diverging lens of focal length $100 \mathrm{mm}$.

Calculate the distance of the image from the lens by using the lens formula.

The diverging lens of focal length $100 \mathrm{mm}$ is replaced by a converging lens also of focal length $100 \mathrm{mm}$. The object is placed $150 \mathrm{mm}$ away from the lens and an image is formed by the converging lens. Compare two properties of this image with those of the image formed by the diverging lens.