Question

(a) Explain what is meant by a virtual, magnified image.
(b) Draw a ray diagram to show the formation of a virtual magnified image of an object by a convex lens. In your diagram, the position of object and image with respect to the principal focus should be shown clearly.
(c) Three convex lenses are available having focal lengths of 4cm, 40cm and 4m respectively. Which one would choose as a magnifying glass and why?

Answer 1

A virtual image is the image that cannot be obtained on a screen. The virtual image is formed when the object is placed between the optical centre and the focus of the lens. Find the magnification in terms of focal length.

Formula used:
$m=\dfrac{v}{u}$
where $v$ and $u$ are the positions of the image and the object with respect to the lens.
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
where $f$ is the focal length of the lens.

Complete step by step answer:
(a) A virtual image is an image that cannot be obtained on a screen. For a mirror, the virtual image of the object is formed inside the mirror. For a lens, the virtual image is formed on the same side of the lens where the rays of light are incident. If the size of a virtual image is larger than the size of the object, then it is called a virtual, magnified image.

(b) The diagram below shows the diagram of a virtual, magnified image of an object placed in front of a convex lens.

The virtual image is formed when the object is placed between the optical centre and the focus of the lens.

(c) The magnification due to a lens is equal to $m=\dfrac{v}{u}$,
where v and u are the positions of the image and the object with respect to the lens.
$\Rightarrow v=um$
From lens maker formula, we get that $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$,
where f is the focal length of the lens.
Substitute the value of v.
$\Rightarrow \dfrac{1}{um}-\dfrac{1}{u}=\dfrac{1}{f}$
$\Rightarrow \dfrac{1}{um}=\dfrac{1}{f}+\dfrac{1}{u}$
$\therefore \dfrac{1}{m}=\dfrac{u}{f}+1$
The value of u is always negative. Therefore, for a constant u, if the value of f decreases, then the value of $\dfrac{u}{f}$ is more negative. As a result, $\dfrac{u}{f}+1$ will be less positive.Therefore, the value of $\dfrac{1}{m}$ decreases. This means that the value m increases. This means that the magnification is very large when the value of the focal length of the converging lens should be very small.

Therefore, the lens of focal length 4cm should be chosen for making the magnifying glass.

Note: When the object is placed beyond the focal length of the lens, the image formed is real and inverted. The virtual image produced by a real object is always erect and the real image produced by a real object is always inverted.