Respuesta :
Answer:
15.1 m
Explanation:
We first calculate the apparent depth from
refractive index, n = real depth/apparent depth
apparent depth, a = real depth/refractive index
real depth = 2.0 m, refractive index = 1.4
apparent depth, a = 2.0/1.4 = 1.43 m
Since the cylindrical cistern has a diameter of 2.9 m, its radius is 2.9/2 = 1.45 m
The angle of refraction, r is thus gotten from the ratio
tan r = radius/apparent depth = 1.45/1.43 = 1.014
r = tan⁻¹(1.014) = 45.4°
The angle of incidence, i is gotten from n = sin i/sin r
sin i = nsin r = 1.4sin45.4° = 1.4 × 0.7120 = 0.9968
i = sin⁻¹(0.9968) = 85.44°
Since the girl's eyes are 1.2 m from the ground, the distance ,h from the edge of the cistern she must stand is gotten from
tan i = h/1.2
h = 1.2tan i = 1.2tan85.44° = 1.2 × 12.54 = 15.05 m
h = 15.05 m ≅ 15.1 m
So, she must stand 15.1 m away from the edge of the cistern to still see the object.
The distance o the object is 3 m far from the girl's eyes who is stan still on the edge of the cylindrical cistern.
What is apparent depth?
It is the depth of the object inside a denser medium when seen from the less dense medium.
[tex]n = \dfrac {D_r}{D_a}[/tex]
Where,
[tex]n[/tex] - refractive index = 1.4
[tex]D_r[/tex] - Real depth = 2 m
[tex]D_a[/tex] - Apparent depth
Put the values in the formula,
[tex]D_a = \dfrac{2} {1.4} \\\\D_a = 1.43[/tex] m
Now from the Pythagorean theorum,
[tex]d = \sqrt{\rm h ^2 + b^2}[/tex]
Where,
[tex]d[/tex] - Apparent distance of the object from observers eye
[tex]h[/tex] - apparent depth from observers eye = 1.43+1.2 = 2.63 m
[tex]b[/tex] - radius of the cylindrical cistern = 1.45 m
Put the values in the formula,
[tex]d = \sqrt{\rm (2.63) ^2 + (1.45 )^2}\\\\d = 3.00 \rm m[/tex]
Therefore, the distance o the object is 3 m far from the girl's eyes who is stan still on the edge of the cylindrical cistern.
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