Question

A section line AB appears to be 10.16 cm on a photograph for which the focal length is 16 cm. The corresponding line measures 2.54 cm on a map, which is to a scale \(\frac{1}{{50,000}}\). The terrain has an average elevation of 200 m above mean sea level. The flying altitude of the aircraft above mean sea level during photograph will be
1. 1800 m
2. 2000 m
3. 2200 m
4. 2400 m

Answer 1

Correct Answer - Option 3 : 2200 m

Concept:

Scale of the map is given by,

\({S_{map}} = \frac{{Map\;distance}}{{Ground\;distance}}\)

Scale of the photo (S) is given by, 

\({\rm{S}} = {\rm{}}\frac{{{\rm{Photo\;distance}}}}{{{\rm{Ground\;distance}}}} = {\rm{}}\frac{{\rm{f}}}{{{\rm{H}} - {{\rm{H}}_{\rm{a}}}}}\)

Calculation:

Given, Focal length, f = 16 cm

Photo distance= 10.16 cm

Map distance = 2.54 cm

Average elevation of terrain, H_a = 200 m

Let height of air craft is H m.

Scale, S_map = 1/50,000

We know that,

\({S_{map}} = \frac{{Map\;distance}}{{Ground\;distance}}\)

\(\frac{1}{{50000}} = \frac{{2.54}}{{Ground\;distance}}\)

Ground distance = 127000 cm = 1270 m

Also

Scale of the photo is 

\({\rm{S_{photo}}} = {\rm{}}\frac{{{\rm{Photo\;distance}}}}{{{\rm{Ground\;distance}}}} = {\rm{}}\frac{{\rm{f}}}{{{\rm{H}} - {{\rm{H}}_{\rm{a}}}}}\)

\({\rm{\;}}{{\rm{S}}_{{\rm{photo}}}} = {\rm{}}\frac{{10.16}}{{127000}} = {\rm{}}\frac{{0.16}}{{{\rm{H}} - 200}}\)

On solving, we get H = 2200 m