# The diameter of Mars is 6794 \ km, and its minimum distance from the earth is 5.58*10^7 \ km....

## Question:

The diameter of Mars is {eq}6794 \ km {/eq}, and its minimum distance from the earth is {eq}5.58*10^7 \ km {/eq}.

When the Mars is at this distance, find the diameter of the image of Mars formed by a spherical, concave, telescope mirror with a focal length of {eq}1.95 \ m {/eq}.

## Image Formed by the Concave Mirror

For a concave mirror position of the image formed can be obtained from the equation {eq}\dfrac { 1 } { f } = \dfrac { 1 } { u } + \dfrac { 1 } { v } {/eq}. Here f, u and v are the focal length of the mirror, object distance from the pole of the mirror, image distance from the pole of the mirror respectively. If {eq}h, \ \ h_1 {/eq} are the object size and image size the their ratio can be expressed as {eq}\dfrac { h_1 }{ h } = \dfrac { v } { u } {/eq}. Using this relation size of the image formed can be obtained.

Given points

• Length of the diameter of mars {eq}h = 6.794 \times 10^6 \ m {/eq}
• Minimum distance of Mars from Earth {eq}u = 5.58 \times 10^{10} \ m {/eq}
• Focal length of the concave mirror in the telescope f = 1.95 m

Let v be the distance from the pole of the mirror where the image will be formed

Then we have the relation {eq}\dfrac { 1 } { f } = \dfrac { 1 } { u } + \dfrac { 1 } { v } {/eq}

So we get {eq}\dfrac { 1 } { v } = \dfrac { 1 } { f } - \dfrac { 1 } { u } \\ \dfrac { 1 } { v } = \dfrac { 1 } { - 1.95 } - \dfrac { 1 } { -5.58 \times 10^{10} } \\ \dfrac { 1 } { v } = - 0.513 {/eq}

The negative sign is given to f and u according to the sign convention.

So the distance at which the image is formed {eq}v = - 1.95 \ m {/eq}

Let {eq}h_1 {/eq} be the size of the image

The size of the image can be obtained from the ratio {eq}\dfrac { h_1 } { h } = \dfrac { v } { u } {/eq}

So size of the image formed {eq}h_1 = h \times \dfrac { v } { u } \\ h_1 = 6.794 \times 10^6 \times \dfrac { 1.95 } { 5.58 \times 10^{10} } \\ h_1 = 2.37 \times 10^{-4} \ m {/eq}