A thin hemispherical bowl of clear plastic floats on water in a tank. The radius

Question

A thin hemispherical bowl of clear plastic floats on water in a tank. The radius of the bowl is 50 cm and the depth of the bowl in water is 10 cm. The depth of the water (n = 1.33) in the tank is h = 590 cm. An object 8.0 cm long is on the bottom of the tank directly below the bowl. The object is viewed from directly above the bowl. Ignore the refractive effects of the plastic. In the figure, the position of the image below the water level, in cm, is closest to:

Choices:

PART 2:

A thin hemispherical bowl of clear plastic floats on water in a tank. The radius of the bowl is 50 cm and the depth of the bowl in water is 10 cm. The depth of the water (n = 1.33) in the tank is h = 360 cm. An object 8.0 cm long is on the bottom of the tank directly below the bowl. The object is viewed from directly above the bowl. Ignore the refractive effects of the plastic. In the figure, the size of the image, in cm, is closest to:

Choices:

Explanation / Answer

given thin hemispherical plastic bowl floating in water

radius of bowl R = 50 cm

depth of bowl in water, d = 10 cm

n = 1.33

h = 5.9 m

H = 8 cm

hence considering this a lens

for object inside water

focal length = f

1/f = (1 - n)(1/R)/n = (1/n - 1)/R

f = -201.5151 cm

for the given situation, u = -(h - d) = -580 cm

v = ?

1/v - 1/u = 1/f

1/v = 1/f + 1/u

v = -149.554 cm

hence from the water level, v = -149.554 - 10 = -159.554 = -160 cm

option e.

2. size of image = 8*(149.554/580) = 2.0689655 cm

option a. 2.9

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