The Sun Is Too Big!

Classical Mechanics Level 5

An amateur astronomer wishes to estimate roughly the size of the sun using his crude telescope consisting of an objective lens of focal length $200\text{ cm}$ and an eyepiece of focal length $10\text { cm}.$ By adjusting the distance of the eyepiece from the objective, he obtains an image of the sun on a screen $40\text{ cm}$ behind the eyepiece.The diameter of the sun's image is measured to be $6.0\text{ cm}.$ Now, the question is what is the estimate of the sun's size, give that the average Earth-sun distance is $1.5 \times 10^{11}\text{ m}.$

Details and Assumptions :

Enter your answer in scientific notation. For example $3 \times 10^8$ can be entered as $3E+8.$

The answer is 1.5E+9.

1 solution

Samara Simha Reddy
Apr 29, 2016

$\large \displaystyle \text{Here } f_0 = 200cm, f_e = 10cm, v_e = +40cm \text{(for real)}\\ \large \displaystyle \text{As } \frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} .\\ \large \displaystyle \therefore \frac{1}{u_e} = \frac{1}{v_e} -\frac{1}{v_e} = \frac{1}{40} - \frac{1}{10}\\ \large \displaystyle = \frac{1-4}{40} = \frac{-3}{40}\\ \large \displaystyle \therefore u_e = -\frac{40}{3}.\\ \large \displaystyle \text{Magnification power produced by eye piece is }\\ \large \displaystyle m_e = \frac{v_e}{|u_e|} = \frac{40}{\frac{40}{3}} = 3.\\ \large \displaystyle \therefore \text{Diameter of the image formed by the objective is }\\ \large \displaystyle d = \frac{6}{3} cm = 2 cm.$

If $D$ is the diameter of the sun (in m), then the angle subtended by it on the objective will be,

$\large \displaystyle \alpha = \frac{D}{1.5 \times 10^{11}} rad$

Angle subtended by the image at the objective will be equal to this angle and is given by,

$\large \displaystyle \alpha = \frac{\text{size of image}}{f_0} = \frac{2}{200} = \frac{1}{100}rad.\\ \large \displaystyle \therefore \frac{D}{1.5 \times 10^{11}} = \frac{1}{100}\\ \large \displaystyle \implies D = \frac{1.5 \times 10^{11}}{100} = \color{#D61F06}{\boxed{1.5 \times 10^9m}}.$

The answer to this problem is slightly incorrect: 1.35E+9 and not 1.5E+9.

The errors occurred computating $\alpha$ , that is in more precise terms:

$\alpha=2\arctan\space\frac{D}{2d}$

The solution is very simply whether one consider that the Sun's image crosses two lens and than has two magnifications, of which the disk on the screen is the result. So

$\large{ D\times\frac{2}{2-d}\times\frac{0.1}{0.4-0.1}=0.06}$

$\large{D=0.06\times\large{\frac{1.5x10^{11}}{2}}\times 3=0.09\times1.510^{11}=1.35\times10^9}$

     Ciao from
             an Old Dude :)

Luciano Riosa - 5 years, 1 month ago

The magnification of such a system is given by (f.l. objective/f.l. eyepiece), that is 20 not 3.

Luciano Riosa - 5 years, 1 month ago

I have given in detail that it is magnification for the eyepiece. Study properly dude.

Samara Simha Reddy - 5 years, 1 month ago

I got 1.4E+9 and it is the real diameter of Sun. I think you got some wrong calculations...

Pushan Paul - 5 years, 1 month ago

The Sun Is Too Big!

The answer is 1.5E+9.

1 solution

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