Calculating Diameter of Illumination on the Moon | Help with Light Problem

[shaiqbashir]

Please help me in this light problem!

HI!

well would u please help me in solving this problem the peoblem is:

"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 I am 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

Thanks in advance

 

[xanthym]

"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 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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[thepatientmental]

.

Hi there!

Yes, the method you have used is correct. The formula you have used is known as the Rayleigh criterion, which is used to calculate the diameter of the illuminated area on a distant object.

To explain how we can take angle as lambda/2d, we need to understand how light spreads out as it travels through space. When light from a point source, such as a laser beam, travels through space, it spreads out in a cone shape. The angle of this cone is determined by the wavelength (lambda) and the size of the source (d). So, for a smaller source size (d), the angle of the cone will be larger, and for a larger source size (d), the angle of the cone will be smaller.

In this case, the angle of the cone is very small because the source (laser beam) and the distance (240000 miles) are both very large compared to the wavelength (6328A). Therefore, we can approximate the angle as lambda/2d, which is the same as the Rayleigh criterion formula you have used.

I hope this helps to clarify things for you. Let me know if you have any other questions. Good luck with your problem!