Edits to “Statement”, “Solution”, “Answer”

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revision #19509 parent #19508 ← older
en/13.2.2.md +14 −10
@@ -1,27 +1,31 @@
### Statement
−$13.2.2.$ [Insert the problem statement]
+$13.2.2.$ a) The apparent depth of a reservoir, when looking vertically downward, is $3$ m. What is its true depth?
+b) An airplane flies over a submarine submerged at a small depth, at an altitude of $3$ km. What will the altitude of the airplane appear to be when observed from the submarine?
+
### Solution
−a) Consider two rays: one goes vertically upward from a point on the bottom, A, so it does not refract and intersects the water surface at point O; the other goes at a small angle β to the vertical, intersects the water surface at point E, and emerges at an angle α.
+a) Consider two rays: one goes vertically upward from a point on the bottom $A$, so it does not refract and intersects the water surface at point $O$; the other goes at a small angle $\beta$ to the vertical, intersects the water surface at point $E$, and emerges at an angle $\alpha$.
The apparent bottom will be at the intersection of the vertical line and the emerging ray.
−From Snell's law, using sin(x) ≈ tan(x) ≈ x:
+From Snell's law, using $\sin(x) \approx \tan(x) \approx x$:
−α = n β
+$$\alpha = n \beta$$
−Express the segment EO in two triangles:
+Express the segment $EO$ in two triangles:
−EO = h' α = h β
+$$EO = h' \alpha = h \beta$$
−h = n h' = 4 m
+$$h = n h' = 4 \text{ m}$$
−b) The solution is analogous to part (a), except here the extension of the refracted ray will go beyond the true bottom, so the apparent depth will be greater:
+b) The solution is analogous to part (a), except here the extension of the refracted ray will go beyond the true bottom, and the apparent depth will be greater:
−h' = h n = 4 km
+$$h' = h n = 4 \text{ km}$$
#### Answer
−[Insert a concise answer or boxed result]
+a) $$h = n h' = 4 \text{ m}$$
+
+b) $$h' = h n = 4 \text{ km}$$