Question

An object 0.600cm tall is placed 16.5cm to the left of the vertex of a convex spherical mirror having a radius of curvature of 22.0cm

- Determine the position of the image.

- Determine the size of the image.

- Determine the orientation of the image.

Answer 1

Concepts and reason

The concepts used to solve this problem are radius of curvature, equation of mirror, focal length, and magnification.

Initially, use the concept of radius of curvature to calculate the focal length of the mirror.

Then, use the concept of mirror equation to calculate the image position.

Then, use the relation between the magnification, image distance, and object distance to calculate the magnification.

Finally, use the concept of sign convention of spherical mirror to find the nature of the formed image.

Fundamentals

Express the relation between the radius of curvature and focal length.

$f = \frac{R}{2}$

Here, the focal length is $f$ and radius of curvature is $R$ .

The expression for the mirror equation is given below:

$\frac{1}{f} = \frac{1}{s} + \frac{1}{{s'}}$

Here, the focal length is $f$ , object distance is $s$ , and image distance is $s'$ .

Express the relation between the image distance and object distance from the magnification of the mirror equation as follows:

$M = - \frac{{s'}}{s}$

Here, the magnification is $M$ , object distance is $s$ , and the image distance is $s'$ .

Express the relation between the height of the object and height of the image from the magnification of the mirror equation as follows:

$M = \frac{{{h_i}}}{{{h_o}}}$

Here, the magnification is $M$ , height of the image is ${h_i}$ , and the height of the object is ${h_o}$ .

The focal length of the mirror is equal to the half of the radius of curvature of the mirror.

The expression for the focal length is as follows:

$f = \frac{R}{2}$

Substitute $22.0\,{\rm{cm}}$ for $R$ .

$\begin{array}{c}\\f = \frac{{22.0\,{\rm{cm }}}}{2}\\\\ = 11.0\,{\rm{cm}}\\\end{array}$

The expression for the mirror equation is given below:

$\frac{1}{f} = \frac{1}{s} + \frac{1}{{s'}}$

Rearrange the above equation for $s'$ .

$\begin{array}{c}\\\frac{1}{{s'}} = \frac{1}{f} - \frac{1}{s}\\\\s' = \frac{{fs}}{{s - f}}\\\end{array}$

Substitute $- 11.0\,{\rm{cm}}$ for $f$ and $16.5\,{\rm{cm}}$ for $s$ .

$\begin{array}{c}\\s' = \frac{{\left( { - 11.0\,{\rm{cm}}} \right)\left( {16.5\,{\rm{cm}}} \right)}}{{\left( {16.5\,{\rm{cm}}} \right) - \left( { - 11.0\,{\rm{cm}}} \right)}}\\\\ = - 6.6\,{\rm{cm}}\\\end{array}$

The ratio of the image distance to the object distance is known as the magnification of the mirror.

The expression for the magnification is as follows:

$M = - \frac{{s'}}{s}$

Substitute $16.5\,{\rm{cm}}$ for $s$ and $- 6.6\,{\rm{cm}}$ for $s'$ .

$\begin{array}{c}\\M = - \frac{{\left( { - 6.6\,{\rm{cm}}} \right)}}{{16.5\,{\rm{cm}}}}\\\\ = 0.4\\\end{array}$

The expression for the magnification using the height of the object and the height of the image is as follows:

$M = \frac{{{h_i}}}{{{h_o}}}$

Substitute $0.4$ for $M$ and $0.600\,{\rm{cm}}$ for ${h_o}$ .

$\begin{array}{c}\\0.4 = \frac{{{h_i}}}{{0.60\,{\rm{cm}}}}\\\\{h_i} = \left( {0.60\,{\rm{cm}}} \right)\left( {0.4} \right)\\\\ = 0.24\,{\rm{cm}}\\\end{array}$

The sign convention of a mirror gives the nature of the image.

For the positive object distance, the object is in front of the mirror and is called as a real object.

For the negative object distance, the object is behind the mirror and is called as a virtual object.

For the positive image distance, the image is in front of the mirror and is called as a real image.

For the negative image distance, the image is behind the mirror and is called as a virtual image.

For the positive magnification, the image is upright with respect to object.

For the negative magnification, the image is inverted with respect to object.

The image distance is negative; therefore, the image is virtual image.

The magnification is positive; therefore, the image is upright.

Thus, the image is virtual and upright.

Ans:

The image is formed at a distance of $6.6\,cm$ on the other side of the mirror.

The height of the image is $0.24\,cm$ .

The image formed by the mirror is virtual and upright.