According to New Cartesian Sign Convention, the height of a virtual image is sometimes negative.

The mirror formula gives the relationship between the object distance, the image distance and the focal length in case of spherical mirrors. It is mathematically represented as:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

Where, $u=$object distance, $v=$ image distance and $f=$focal length

The image produced by a spherical mirror can be larger than the object, smaller than the object or of the same size as the object depending upon the position of the object and the type of spherical mirror. The size of the image relative to the object is given by the linear magnification, $m$ given by the formula:

$m=\frac{h\text{'}}{h}$

Where, $h$ is the object height and $h\text{'}$ is the image height.

Now consider the following situation:

A $4 \mathrm{cm}$-long object is placed at a distance of $16 \mathrm{cm}$ in front of a concave mirror. The image formed is $6 \mathrm{cm}$ long and is real.

On the basis of the above information and what you have learnt in the course, answer the following question:

What is the magnification produced?

The mirror formula gives the relationship between the object distance, the image distance and the focal length in case of spherical mirrors. It is mathematically represented as:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Where, $u=$object distance, $v=$ image distance and $f=$focal length
The image produced by a spherical mirror can be larger than the object, smaller than the object or of the same size as the object depending upon the position of the object and the type of spherical mirror. The size of the image relative to the object is given by the linear magnification, $m$ given by the formula:
$m=\frac{h\text{'}}{h}$
Where, $h$ is the object height and $h\text{'}$ is the image height.
Now consider the following situation:
A $4 \mathrm{cm}$-long object is placed at a distance of $16 \mathrm{cm}$ in front of a concave mirror. The image formed is $6 \mathrm{cm}$ long and is real.
On the basis of the above information and what you have learnt in the course, answer the following question:

Where is the image formed?

The mirror formula gives the relationship between the object distance, the image distance and the focal length in case of spherical mirrors. It is mathematically represented as:

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

Where, $u=$object distance, $v=$ image distance and $f=$focal length

The image produced by a spherical mirror can be larger than the object, smaller than the object or of the same size as the object depending upon the position of the object and the type of spherical mirror. The size of the image relative to the object is given by the linear magnification, $m$ given by the formula:

$m=\frac{h\text{'}}{h}$

Where, $h$ is the object height and $h\text{'}$ is the image height.

Now consider the following situation:

A $4 \mathrm{cm}$-long object is placed at a distance of $16 \mathrm{cm}$ in front of a concave mirror. The image formed is $6 \mathrm{cm}$ long and is real.

On the basis of the above information and what you have learnt in the course, answer the following question:

What is the focal length of the mirror?

Match the applications of mirrors given in Column I to the type of mirrors given in Column II.

For a concave mirror of focal length $15 \mathrm{cm}$, the description of images produced are mentioned in Column I. Match these descriptions to the object distances mentioned in Column II.