A concave spherical mirror has a radius of curvature of magnitude 19.2 cm. a. Find the location...

Question:

A concave spherical mirror has a radius of curvature of magnitude 19.2 cm.

a. Find the location of the image for the object placed at distances 39.2 cm, 19.2 cm and 9.6 cm.

b. For each case mentioned in part a state whether the image is real or virtual.

c. For each case mentioned in part a state whether the image is upright or inverted.

d. For each case mentioned in part a find the magnification of the image.

Concave mirror:

A spherical mirror is formed by polishing one side of the part of a sphere of glass. If the side opposite to the cavity of the part of the sphere is polished, the mirror formed will be concave, and if the cavity of the sphere part itself is polished, then the mirror formed will be a convex mirror. The focal length of the concave mirror is taken positive.

Given data:

The radius of curvature of the concave spherical mirror is, {eq}R = + 19.2\;{\rm{cm}} {/eq}

Thud the focal length of the mirror will be,

{eq}\begin{align*} f &= \dfrac{R}{2}\\ f &= \dfrac{{19.2}}{2}\\ f &= 9.6\;{\rm{cm}} \end{align*} {/eq}

The expression for the mirror formula is,

{eq}\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} - - - - - - - - \left( 1 \right) {/eq}

a)

The location of the object placed at a distance,

{eq}u = 39.2\;{\rm{cm}} {/eq}

Thus by putting the values in above equation (1)

{eq}\begin{align*} \dfrac{1}{v} + \dfrac{1}{{39.2}} &= \dfrac{1}{{9.6}}\\ \dfrac{1}{v} &= \dfrac{1}{{9.6}} - \dfrac{1}{{39.2}}\\ \dfrac{1}{v} &= \dfrac{{39.2 - 9.6}}{{376.32}}\\ \dfrac{1}{v} &= \dfrac{{29.6}}{{376.32}}\\ \dfrac{1}{v} &= \dfrac{1}{{0.0786}}\\ v &= 12.714\;{\rm{cm}} \end{align*} {/eq}

The location of the object placed at a distance,

{eq}u = 19.2\;{\rm{cm}} {/eq}

Thus by putting the values in above equation (1)

{eq}\begin{align*} \dfrac{1}{v} + \dfrac{1}{{19.2}} &= \dfrac{1}{{9.6}}\\ \dfrac{1}{v} &= \dfrac{1}{{9.6}} - \dfrac{1}{{19.2}}\\ \dfrac{1}{v} &= \dfrac{{19.2 - 9.6}}{{184.32}}\\ \dfrac{1}{v} &= \dfrac{{9.6}}{{184.32}}\\ \dfrac{1}{v} &= \dfrac{1}{{0.052}}\\ v &= 19.2\;{\rm{cm}} \end{align*} {/eq}

b)

The location of the object placed at a distance,

{eq}u = 9.6\;{\rm{cm}} {/eq}

Thus by putting the values in above equation (1)

{eq}\begin{align*} \dfrac{1}{v} + \dfrac{1}{{9.6}} &= \dfrac{1}{{9.6}}\\ \dfrac{1}{v} &= \dfrac{1}{{9.6}} - \dfrac{1}{{9.6}}\\ \dfrac{1}{v} &= 0\\ v \to \infty \end{align*} {/eq}

b)

When the location of the image is comes out to be positive, it can be said that the image is real that is on the same side of the mirror where the object is placed.

For {eq}u = 39.2\;{\rm{cm}} {/eq} image is real

For {eq}u = 19.2\;{\rm{cm}} {/eq} image is real

For {eq}u = 9.6\;{\rm{cm}} {/eq} image cannot be traced out

c)

As the image formed with the mirror real it will always be inverted, and if it is imaginary it will be erect.

For {eq}u = 39.2\;{\rm{cm}} {/eq} image is inverted.

For {eq}u = 19.2\;{\rm{cm}} {/eq} image is inverted.

For {eq}u = 9.6\;{\rm{cm}} {/eq} image view cannot be traced out.

d)

The expression for the magnification is,

{eq}m = \dfrac{{ - v}}{u} {/eq}

Thus,

For {eq}u = 39.2\;{\rm{cm}} {/eq},

{eq}\begin{align*} m &= \dfrac{{ - 12.714}}{{39.2}}\\ m &= - 0.324 \end{align*} {/eq}

For {eq}u = 19.2\;{\rm{cm}} {/eq},

{eq}\begin{align*} m &= \dfrac{{ - 19.2}}{{19.2}}\\ m &= - 1 \end{align*} {/eq}

For, {eq}u = 9.6\;{\rm{cm}} {/eq} ,

{eq}m \to \infty {/eq}