Question

The sun is 150,000,000 km from earth; its diameter is 1,400,000 km.For a science project on solar power, a student uses a 24-cm-diameter converging mirror with a focal length of 47 cm to focus sunlight onto an object. This casts an image of the sun on the object. For the most intense heat, the image of the sun should be in focus.

a. Where should the object be placed?

b. What is the diameter of the image?

c. The intensity of the incoming sunlight is 1050 W/m2. What is the total power of the light captured by the mirror?

d. What is the intensity of sunlight in the projected image? Assume that all of the light captured by the mirror is focused into the image.

Please explain every part neatly and clearly, I am having a very difficult time finding a clear answer and process for this on chegg(or on my own). Thank you!

Answer 1

a)

d_o = distance of the sun = 1.5 x 10¹¹ m

d_i = distance of the object = ?

f = focal length of the converging mirror = 47 cm = 0.47 m

Using the mirror equation

1/d_i + 1/d_o = 1/f

1/d_i + 1/(1.5 x 10¹¹) = 1/0.47

d_i = 0.47 m

b)

h_o = height of the sun = diameter of sun = 1.4 x 10⁹ m

h_i = height of image = ?

Using the magnification equation

h_i /h_o = - d_i/d_o

h_i /(1.4 x 10⁹) = - (0.47)/(1.5 x 10¹¹)

h_i = - 4.4 x 10^-3 m

so diameter of image = 4.4 x 10^-3 m

c)

I = intensity of incoming light = 1050 W/m²

d = diameter of mirror = 24 cm = 0.24 m

A = area of mirror = (0.25)d² = (0.25) (3.14) (0.24)² = 0.045216 m²

Total power of the light captured is given as

P = I A

P = (1050) (0.045216)

P = 47.5 Watt

d)

D_i = diameter of image = 4.4 x 10^-3 m

A_i = area of the image = (0.25) D_i² =  (0.25) (3.14) (4.4 x 10^-3)² = 1.52 x 10^-5 m²

I_i = intensity of light at the image

I_i = P /A_i

I_i = 47.5/(1.52 x 10^-5)

I_i = 3.125 x 10⁶ W/m²