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Question: In aerial mapping a camera uses a lens with a \[100{\rm{ cm}}\] focal length. The height at which th...

In aerial mapping a camera uses a lens with a 100cm100{\rm{ cm}} focal length. The height at which the airplane must fly, so that the photograph of a 1km1{\rm{ km}} long strip on the ground fits exactly on the 20cm20{\rm{ cm}} long filmstrip of the camera, is
A. 200km200{\rm{ km}}
B. 20km20{\rm{ km}}
C. 5km5{\rm{ km}}
D. 1km1{\rm{ km}}

Explanation

Solution

In the solution, we will use the expression for lens formula which gives us the relation between the focal length of the lens, the distance of strip from the lens and distance of the image from the lens. We will also use the expression of the relation between image distance, strip distance, length of the strip and length of the image formed.

Complete step by step answer:
Given:
The actual length of the strip is l1=1km×(1000m1km)=1000m{l_1} = 1{\rm{ km}} \times \left( {\dfrac{{1000{\rm{ m}}}}{{{\rm{1 km}}}}} \right) = 1000{\rm{ m}}.
The length of the image formed is l2=20cm×(1m100cm)=0.2m{l_2} = 20{\rm{ cm}} \times \left( {\dfrac{{1{\rm{ m}}}}{{{\rm{100 cm}}}}} \right) = 0.2{\rm{ m}}
The focal length of the camera lens is f=100cm×(1m100cm)=1mf = 100{\rm{ cm}} \times \left( {\dfrac{{1{\rm{ m}}}}{{{\rm{100 cm}}}}} \right) = 1{\rm{ m}}.

We have to find the height at which the airplane must fly so that the photograph of the strip fits exactly on 20cm20{\rm{ cm}} long filmstrip of the camera.

The relation between image distance and strip distance is expressed as below:
vu=−l2l1\dfrac{v}{u} = - \dfrac{{{l_2}}}{{{l_1}}}……(1)
Here, v is the image distance from the lens and u is the distance of the image from the camera.

Substitute 0.2m0.2{\rm{ m}} for l2{l_2} and 1000m1000{\rm{ m}} for l1{l_1} in equation (1).

\dfrac{v}{u} = - \dfrac{{0.2{\rm{ m}}}}{{1000{\rm{ m}}}}\\\ v = - 2 \times {10^{ - 4}}u \end{array}$$ Write the expression lens formula. $$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$$ Substitute $$1{\rm{ m}}$$ for f and $$ - 2 \times {10^{ - 4}}u$$ for v in the above expression. $$\begin{array}{l} \dfrac{1}{{1{\rm{ m}}}} = \dfrac{1}{{\left( { - 2 \times {{10}^{ - 4}}u} \right)}} - \dfrac{1}{u}\\\ u = - 5000{\rm{ m}} \times \left( {\dfrac{{{\rm{km}}}}{{{\rm{1000 m}}}}} \right)\\\ u = 5{\rm{ km}} \end{array}$$ Therefore, $$5{\rm{ km}}$$ is the height at which the airplane must fly in order to fit the strip in the given stip of the camera film. **So, the correct answer is “Option C”.** **Note:** Do not forget to convert the given values into the same units before substituting them into the respective equations. Also, we have to take care of the negative sign present in equation (1). It will be an added advantage if we remember the conversion of various units of length.