Use the law of reflection and index of refraction to find apparent d

[rocapp]

Homework Statement

See image

Homework Equations

n1sin(theta1) = n2sin(theta2)
1/s + 1/s' = 1/f

The Attempt at a Solution

My question is for the second part. I think I've figured out why I got it wrong, but I'm not sure.

So since the object distance (s) should be approximately 21 cm, then you apply the refraction theory to find

s'=s*n1/n2
s'=21*(1.33)
s'=~28 cm

Is this correct?

 

[Doc Al]

No, not correct. How did you do the first part, since it's the same kind of problem? For the second part, where is the image of the fish in the mirror? That's your "object".

Once you have the correct depth of the "object", do this. Imagine a small cone of light coming from that object extending up towards the surface. If that cone has an angle θ, what will be the angle of the cone after it leaves the water? Use that and a bit of trig to figure out the apparent depth of the object. (Assume the angles involved are small.)

 

[rocapp]

For the first part, I used

(Na/Sa) + (Nb/Sb) = (Nb - Na) / R

1.3/6 + 1/x = nb - na / R

.217 + 1/x = nb - na / infinity

1/x = .217

x = 4.61 cm

 

[Doc Al]

For the first part, I used

(Na/Sa) + (Nb/Sb) = (Nb - Na) / R

1.3/6 + 1/x = nb - na / R

.217 + 1/x = nb - na / infinity

1/x = .217

x = 4.61 cm

That's an excellent way to solve the problem. But don't round off until the end and you'll get the correct answer.

The second part can be handled the same way, once you find the correct image location.

 

[Nickie]

Yes, your approach is correct. The second part of the question asks for the apparent distance (d) of the object, which can be found using the formula:

d = s + s'

where s is the object distance and s' is the image distance. As you correctly calculated, the image distance is 28 cm. Therefore, the apparent distance of the object is:

d = 21 cm + 28 cm = 49 cm

This means that the object appears to be 49 cm away from the observer due to the refraction of light through the medium with an index of refraction of 1.33.