Question 2

2A) A diver at a depth of $100~{\rm m}$ below the surface of the water in the Puget Sound looks up at an angle of 45^o to the normal and sees the top of the Space Needle. Assume that the top of the Space Needle is $200~{\rm m}$ above sea level. The water has refractive index n = 1.33, and the water in the Puget Sound is unusually calm (i.e. the surface of the water is flat).

i) [15 pts] What is the horizontal distance between the diver and the Space Needle?

The horizontal distance between the diver and the point at which the light from the top Space Needle that the diver sees enters the water is

$$d_2~=~100\tan{\pi\over 4}~=~100~{\rm m}$$

Snells law $n_1\sin\theta_1~=~n_2\sin\theta_2$tells us that the refracted light is at an angle of $\sin\theta_1~=~n\sin{\pi\over 4}$, giving $\theta_1~=~70.1^o$. Thus, we have that the horizontal distance between the point at which the light from the top Space Needle that the diver sees enters the water, and the Space Needle is

$$d_1~=~200\tan\theta_1~=~553.3~{\rm m}$$

and hence the total horizontal distance between the diver and the Space Needle is

$$d_{\rm tot}~=~d_1+d_2~=~553.3+100~=~653.3~{\rm m}$$

ii)[5 pts] At which angle relative to the normal will the diver only see people lying on the beach? (Light reflected from their bodies will be incident on the water at $\sim~90^o$.)

For the diver only to see people on the beach means that $\theta_1~=~90^o$, and hence the diver would need to look at

$$\sin\theta_c~=~{1\over n}~=~48.75^o$$

2B) [7 pts] Two point sources of $500~{\rm nm}$ light are separated by a distance $1~{\rm m}$. Given that the aperture in your eye has a diameter of $a~=~2~{\rm mm}$, at what maximum distance between you and the point sources can you determine that there are two sources of light and not just one?

Rayleighs criterion tells us that the angle at which we can no longer resolve two point source due to diffractions is defined by

$$a\sin\theta~=~1.22\lambda$$

For two distant objects the angle they subtend a distance d away is

$$\theta\sim {h\over d}$$

where h is the seperation between the sources, and hence (using the small angle limit of $\sin\theta\sim\theta$), we have that in order to resolve the two sources

$$d\le{ a h\over 1.22\lambda}~=~3.28~{\rm km}$$

2C) [6 pts] A long wire of diameter $1200~{\rm nm}$ is placed in the path of a laser beam of wavelength $600~{\rm nm}$ and diameter $1~{\rm cm}$. Describe and explain the intensity pattern observed on a screen placed $10~{\rm m}$from the wire, indicating intensity minima and maxima.

Babinetts Principle tells us that the diffraction pattern away from the beam is exactly the same at that produced by a slit of the same width. Diffraction minima are located at angles determined by

$$a\sin\theta~=~p\lambda$$

where $p~=~0, \pm1 , \pm 2, ...$ is an integer. The complete intensity pattern, away from the forward direction is given by

$$I(\theta)~=~I_0\left({\sin\alpha\over\alpha}\right)$$

where

$$\alpha~=~{\pi a\sin\theta\over\lambda}$$

Now given that $a~=~1200~{\rm nm}$ and $\lambda~=~600~{\rm nm}$, we have that

$$\alpha~=~2\pi\sin\theta$$

and so there is a diffraction minimum at $\sin\theta~=~\pm {1\over 2}$, and one at $\theta~=~\pm 90^o$.