Physics Help Lenses (new problem post 1 help please)

[Deathslayer7]

A radar tower sends out a signal of wavelength lambda. It is x meters tall, and it stands on the edge of the ocean. A weather balloon is released from a boat that is a distance d out to sea. The balloon floats up to an altitude h. In this problem, assume that the boat and balloon are so far away from the radar tower that the small angle approximation holds.

Due to interference with reflections off the water, certain wavelengths will be weak when they reach the balloon. What is the maximum wavelength that will interfere destructively?
Express your answer in terms of x, h, and d.

Picture:
http://i198.photobucket.com/albums/aa125/death_slayer7/physics.jpg
where the orange line is the first wavelength, the blue line is the second wavelength which reflects off the water, but also a part goes through.
The black square is the radio tower, and the circle the hot air ballon.


my work:

I extend the second wavelength, so it touched down (h+x) length while the first is (h-x) length.

From there i proceeded to find the length difference and set it equal to lambda/2.

for L1(the orange) i got

L1^2= h^2 + 2*h*x+x^2+d^2

L2^2= h^2 - 2*h*x +x^2 +d^2

take the sqrt of both and subtract L1 from L2 you get this:

L2-L1= sqrt(h^2 + 2*h*x+x^2+d^2) - sqrt(h^2 - 2*h*x +x^2 +d^2)=lambda/2

[Astrella]

I assume it's a diverging lens since the focal length is negative?

Then the object distance is positive when it's a real object (in front of lens) and negative when it's in the back (virtual object).

Image distance is positive when the image is in the back of the lens (real image) and negative when it's in front of the lens (virtual object).

[Deathslayer7]

i assume its diverging as well, while it doesn't explicitly state it, in part A your dealing with a diverging lens and this is part b of the same question.

so in this case the image is in front of the lens, so its negative.

thats(-1/7.5)+(1/3.70)=1/d

d= 7.3 cm

Seems right? :smallconfused:

edit: just put the answer in. It's right ^^ thanks.

and watch out cause i'll be continually updating this thread throughout the day.

[Deathslayer7]

so i found the magnification of the lens to be 0.51 (which is just -(-3.7)/7.3).

Now its asking where should the object be moved to have a larger magnification.

since M=s/d where s is the image distance diveded by the object distance.

since d is in the denominator, it must decrease to make M bigger right?

but if it passes the focal point does that matter?

[Deathslayer7]

A tank whose bottom is a mirror is filled with water to a depth of 20.3cm . A small fish floats motionless 6.70 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?

Express your answer in centimeters. Use 1.33 for the index of refraction of water

Hint:

Keep in mind that the light has to bounce from the bottom of the fish, down to the mirror, and then back up and out of the water to the viewer. First think of how far down the fish would appear if the index of refraction were the same in the water as out. You can use the law of reflection for a plane mirror to determine where the fish would appear without the water. Then apply the correction for the real refraction due to the water.


having trouble imagining it here.

[Deathslayer7]

new problem post 1 if anyone can help. Thanks. :smallsmile:

[littlebottom]

lambda.


thank you so much, my old physics teacher kept calling it lavender and it got to the point i couldnt remember the actual name, :smallbiggrin: thank you so much.

dont ask me why he did it, i recon its because he could never remember what it was called back in school so he just called it lavender instead to help him remember.... he was a strange but cool teacher.

also, i might give this one a go, but its about 2AM here and ive got a really bad cold and head ache and cant talk due to a really soar throat, so my mind is elsewhere at the moment, sorry :smallredface: