Question:
An oblique aerial photograph of a cricket stadium is taken which includes two linear practice pitches (p and q). The length of both the pitches measured on ground is \( 15\ \text{m} \). The dimension of the image is \( 230\ \text{mm} \times 230\ \text{mm} \), while the photo lengths of p and q are \( 5\ \text{mm} \) and \( 15\ \text{mm} \), respectively.
What are the photo scales for p and q, respectively?
An oblique aerial photograph of a cricket stadium is taken which includes two linear practice pitches (p and q). The length of both the pitches measured on ground is \( 15\ \text{m} \). The dimension of the image is \( 230\ \text{mm} \times 230\ \text{mm} \), while the photo lengths of p and q are \( 5\ \text{mm} \) and \( 15\ \text{mm} \), respectively.
What are the photo scales for p and q, respectively?
What are the photo scales for p and q, respectively?
Show Hint
Photo scale for a line equals photo length divided by the true ground length of that line; compute it separately for pitch p and pitch q.
Updated On: Jul 20, 2026
- \( \dfrac{1}{3000} \) and \( \dfrac{1}{1000} \)
- \( \dfrac{1}{1000} \) and \( \dfrac{1}{3000} \)
- \( \dfrac{1}{300} \) and \( \dfrac{1}{100} \)
- \( \dfrac{1}{100} \) and \( \dfrac{1}{300} \)
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The Correct Option is A
Solution and Explanation
Step 1: Recall the definition of photo scale.
The scale of a photograph along any line is the ratio of the distance measured on the photograph to the corresponding distance measured on the ground, \( S = \dfrac{\text{photo length}}{\text{ground length}} \). Both practice pitches p and q have the same true ground length of \( 15\ \text{m} = 15000\ \text{mm} \), but because the photograph is oblique the scale is not uniform across the image, so p and q can have different photo scales even though their ground lengths are equal.
Step 2: Compute the scale for pitch p.
The photo length of p is \( 5\ \text{mm} \).\[ S_p = \frac{5\ \text{mm}}{15000\ \text{mm}} = \frac{1}{3000} \]
Step 3: Compute the scale for pitch q.
The photo length of q is \( 15\ \text{mm} \).\[ S_q = \frac{15\ \text{mm}}{15000\ \text{mm}} = \frac{1}{1000} \]
Step 4: Compare with the given options.
The pair \( \left(\dfrac{1}{3000}, \dfrac{1}{1000}\right) \) matches option (A). Option (B) simply swaps the two scales, and options (C) and (D) both use an incorrect ground length (100 m instead of 15 m) and are swapped versions of each other, so none of them are consistent with the given photo lengths against the 15 m ground length.\[ \boxed{S_p = \tfrac{1}{3000},\ S_q = \tfrac{1}{1000}} \]
The scale of a photograph along any line is the ratio of the distance measured on the photograph to the corresponding distance measured on the ground, \( S = \dfrac{\text{photo length}}{\text{ground length}} \). Both practice pitches p and q have the same true ground length of \( 15\ \text{m} = 15000\ \text{mm} \), but because the photograph is oblique the scale is not uniform across the image, so p and q can have different photo scales even though their ground lengths are equal.
Step 2: Compute the scale for pitch p.
The photo length of p is \( 5\ \text{mm} \).\[ S_p = \frac{5\ \text{mm}}{15000\ \text{mm}} = \frac{1}{3000} \]
Step 3: Compute the scale for pitch q.
The photo length of q is \( 15\ \text{mm} \).\[ S_q = \frac{15\ \text{mm}}{15000\ \text{mm}} = \frac{1}{1000} \]
Step 4: Compare with the given options.
The pair \( \left(\dfrac{1}{3000}, \dfrac{1}{1000}\right) \) matches option (A). Option (B) simply swaps the two scales, and options (C) and (D) both use an incorrect ground length (100 m instead of 15 m) and are swapped versions of each other, so none of them are consistent with the given photo lengths against the 15 m ground length.\[ \boxed{S_p = \tfrac{1}{3000},\ S_q = \tfrac{1}{1000}} \]
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