How Is the Refractive Index Calculated from Apparent and Real Depths?

[SammyD97]

Homework Statement

Hi. I need help with homework. the correct answers have been provided in the question. one doesn't match with mine and the other does. I don't understand how the one answer can be correct when its dependent on my incorrect answer.

A transparent cube of 15cm edge contains a small air bubble. its apparent depth when viewed when viewed from one face of the cube is 6cm and when viewed from the opposite face is 4cm. what is the refractive index of the substance of the cube and what is the actual distance of the bubble from the first face.
Answers: Refractive index=1.5 real depth=9cm

Homework Equations

apparent depth/real depth=n2/n1

The Attempt at a Solution

6/real depth=n2/1
6=n2*real depth
rd=6/n2...(1) where rd=real depth

4/(15-rd)=n2/1
4=n2(15-rd)
4=15n2-n2*rd...(2)

substitute (1) into (2)
4=15n2-n2(6/n2)
4=15n2-6
10=15n2
n2=refractive index=2/3 (I thought the lowest refractive index was that in a vacuum where its 1)

6/real depth=2/3/1
6=2/3*real depth
real depth=9cm

 

[SammyS]

Homework Statement

Hi. I need help with homework. the correct answers have been provided in the question. one doesn't match with mine and the other does. I don't understand how the one answer can be correct when its dependent on my incorrect answer.

A transparent cube of 15cm edge contains a small air bubble. its apparent depth when viewed when viewed from one face of the cube is 6cm and when viewed from the opposite face is 4cm. what is the refractive index of the substance of the cube and what is the actual distance of the bubble from the first face.
Answers: Refractive index=1.5 real depth=9cm

Homework Equations

apparent depth/real depth=n2/n1

The Attempt at a Solution

6/real depth=n2/1
6=n2*real depth
rd=6/n2...(1) where rd=real depth

4/(15-rd)=n2/1
4=n2(15-rd)
4=15n2-n2*rd...(2)

substitute (1) into (2)
4=15n2-n2(6/n2)
4=15n2-6
10=15n2
n2=refractive index=2/3 (I thought the lowest refractive index was that in a vacuum where its 1)

6/real depth=2/3/1
6=2/3*real depth
real depth=9cm

Hello SammyD97. Welcome to PF ! (I thought I should be first to reply considering our user names.)

Your second answer is correct, because you make the same mistake in arriving at both answers, but the effect is to compensate for the mistake made in getting the first one.

Check on what you have done as follows:
Use the correct index of refraction, 1.5, and use your method to predict the apparent distance, knowing that the true distance is 9 cm from a surface.