Question

A vertical photograph of a flat area having an average elevation of 250 m above means sea level was taken with a camera of focal length 25 cm. A section Line AB 300 m long in the area measures 15 cm on the photograph; a tower BP in the area also appears on the photograph. The distance between images of the top and bottom of the tower measures 0.5 cm on the photograph. The distance of the Image of the top of the tower is 10 cm. The actual height of the tower is
1. 10 m
2. 15 m
3. 20 m
4. 25 m

Answer 1

Correct Answer - Option 4 : 25 m

Concept:

Scale of the photograph is given by,

\(S{\rm{cale}} = {\rm{\;}}\frac{{{\rm{Distance\;on\;photo\;}}}}{{{\rm{Distance\;on\;ground\;}}}} = \frac{{\rm{f}}}{{{\rm{H}} - {{\rm{h}}_{\rm{a}}}}}\)

The distance between images of top and bottom of the tower measures on photographs is called relief displacement (d) and it is given by:

\(d = \frac{{rh}}{{{\rm{H}} - {{\rm{h}}_{\rm{a}}}}}\)

ha = Average height from MSL, f = Focal length, H = Height from where the photograph is taken, r =  Distance of the Image of the top of the tower, h = Height of tower

Calculation:

Given, h_a = 250 m, f = 25 cm, The distance on ground = 300 m and distance on photo = 15 cm

We know that

\(S{\rm{cale}} = {\rm{\;}}\frac{{{\rm{Distance\;on\;photo\;}}}}{{{\rm{Distance\;on\;ground\;}}}} = \frac{{\rm{f}}}{{{\rm{H}} - {{\rm{h}}_{\rm{a}}}}}\)

\(\frac{{15{\rm{\;}}}}{{300}} = \frac{{25}}{{{\rm{H}} - 250}}\)

∴ H = 750 m

\(d = \frac{{rh}}{{{\rm{H}} - {{\rm{h}}_{\rm{a}}}}}\)

Where r = 10 cm

\(0.5 = \frac{{10\; \times \;h}}{{{\rm{\;}}750\; - \;250}}\)

∴  h = 25 m