Refraction problem, involving mirrors

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SUMMARY

The discussion focuses on calculating the apparent depth of a fish in a water-filled tank with a mirrored bottom. The fish is located 7.50 cm below the water's surface, and the tank is filled to a depth of 19.3 cm. Using the formula for apparent depth, s' = s/(n1/n2), where n1 is the refractive index of water (1.33), the correct apparent depth of the fish's reflection is determined to be 5.64 cm when viewed at normal incidence. The user also calculated the apparent depth of the mirror to be 14.5 cm, leading to confusion in the final calculation.

PREREQUISITES
  • Understanding of refraction and the concept of apparent depth
  • Familiarity with the refractive index, specifically for water (n1 = 1.33)
  • Knowledge of mirror properties and virtual images
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the principles of refraction in optics, focusing on Snell's Law
  • Learn about the properties of mirrors and virtual images in physics
  • Explore the concept of apparent depth in different mediums
  • Practice solving problems involving refraction and reflection using various scenarios
USEFUL FOR

Students studying physics, particularly those focusing on optics and refraction, as well as educators looking for examples of real-world applications of these concepts.

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Homework Statement



A tank whose bottom is a mirror is filled with water to a depth of 19.3 cm. A small fish floats motionless 7.50 cm under the surface of the water.

What is the apparent depth of the reflection of the fish in the bottom of the tank when viewed at normal incidence?

Homework Equations



s = (n1/n2) * s'
therefore
s' = s/(n1/n2)

The Attempt at a Solution



I found the apparent depth of the fish to be 5.64 (viewed from the normal). Then i found the apparent depth of the mirror to be 14.5, and then used the property of mirrors (virtual image distance = real distance, s=s')

(14.5-5.64)*2=17.72, incorrect
 
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nevermind, i got it

i did [[(19.3*2) -7.5] /1.33]
 

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