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Zorluk: OrtaRefraction of Light, Total Internal Reflection, and Prisms

A coin lies at the bottom of a vessel filled with a liquid to a depth of 14.0 cm14.0\text{ cm}. If the refractive index of the liquid relative to air is 1.401.40, calculate the apparent upward displacement of the coin in centimeters when viewed vertically from directly above.

Cevap: 4 cm

Cevap

The apparent upward displacement of the coin is 4.0 cm4.0\text{ cm}.
Refraction at the liquid-air boundary makes an object at real depth h=14.0 cmh = 14.0\text{ cm} appear at an apparent depth h=hn=14.01.40=10.0 cmh' = \frac{h}{n} = \frac{14.0}{1.40} = 10.0\text{ cm}. The apparent upward displacement is the difference between real depth and apparent depth: d=14.0 cm10.0 cm=4.0 cmd = 14.0\text{ cm} - 10.0\text{ cm} = 4.0\text{ cm}.

Adım Adım Çözüm

1
Identify the given values and formula for refractive index in terms of depth.
Real depth h=14.0 cmh = 14.0\text{ cm}, refractive index n=1.40n = 1.40. Formula: n=Real depthApparent depth=hhn = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{h}{h'}.
Light rays bending away from the normal upon leaving the denser liquid cause the coin to appear closer to the surface.
2
Calculate the apparent depth (hh').
h=14.0 cm1.40=10.0 cmh' = \frac{14.0\text{ cm}}{1.40} = 10.0\text{ cm}.
Rearranging the refractive index formula gives h=hnh' = \frac{h}{n}.
3
Calculate the apparent upward displacement (dd).
d=hh=14.0 cm10.0 cm=4.0 cmd = h - h' = 14.0\text{ cm} - 10.0\text{ cm} = 4.0\text{ cm}.
The displacement is the distance between the actual position at the bottom and the virtual image position.

Anahtar Kavram

Real depth, apparent depth, and apparent displacement
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