An amateur astronomer wishes to estimate roughly the size of the sun using his crude telescope consisting of an objective lens of focal length 200cm and an eyepiece of focal length 10cm. By adjusting the distance of the eye-piece from the objective, he obtains an image of the sum on a screen 40cm behind the eyepiece. the diameter of the sun's image is measured to be 6.0cm. What is the estimate of the sun's size, given that the average earth-sun distance is 1.5×10 11 m.

Question

An amateur astronomer wishes to estimate roughly the size of the sun using his crude telescope consisting of an objective lens of focal length 200cm and an eyepiece of focal length 10cm. By adjusting the distance of the eye-piece from the objective, he obtains an image of the sum on a screen 40cm behind the eyepiece. the diameter of the sun's image is measured to be 6.0cm. What is the estimate of the sun's size, given that the average earth-sun distance is 1.5×10 11 m.

Answer 1

We know that

\(\frac1f=\frac1u-\frac1v\)

\(\frac1u=\frac1v-\frac1f\) 

\(\frac1u=\frac1{40}-\frac1{10}\) 

\(\frac1u=\frac3{40}\) 

\(u=\frac{40}3\)

Magnification power = \(\frac{v}u\) = \(\cfrac{40}{\frac{40}3}\)  = 3

Diameter of magnification formed

d = 6/3

d = 2 cm

if D is the diameter of image formed by the sun, then the angle

\(\alpha=\frac{D}{1.5\times10^{11}m}\) 

Then

\(\alpha=\frac{size\,of\,image}{f_0}\) 

\(\frac{D}{1.5\times 10^{11}} = \frac2{200}\) 

D = \(\frac{1.5\times10^{11}}{100}\) 

D = \(\)1.5 x 10⁹ m