1. Mar 4, 2005

### shaiqbashir

HI!

"a laser beam of diameter 1cm is pointed on the moon. what is the diameter of the area illuminated on the moon. The moon is about 240000 miles away. take wavlength=6328A."

The solution that im using is this.

tan (lambda/2d)=x/2/R ; (R is the distance between the moon and the slit)

since lamda<<d

therefore tan (lambda/2d)=lambda/2d

so we now have:

lambda/2d=x/2R
or x=lamda(R)/d

where x is the required answer.

Now tell me that that is this the right method to get the required answer?

if not then what is the right method?

secondly can u explain how we can take angle as lamda/2d

2. Mar 4, 2005

### xanthym

The laser is diffracted by its CIRCULAR aperture, which has different properties than single-slit diffraction. Fraunhofer diffraction applies in the far-field for our case:
{Angular Displacement of First Minimum} = θ1 :SUCH THAT: sin(θ1) = (1.22)λ/d
where "d" is laser's circular aperture diameter. In the far field:
{Radius of First Minimum on Target} = r = D*sin(θ1)
where D is target distance. From the first equation above and using {D = 240000 mi = 386e(6) meter}, {λ = 6328e(-10) m}, {d = 1.0e(-2) m}:
r = D*(1.22)λ/d = {386e(6) m}*(1.22)*(6328e(-10) m)/(1.0e(-2) m) = (29,800 m)
{Laser Spot Diameter On Moon} = 2*r = (59,600 meters)

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Last edited: Mar 4, 2005