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.

Where should the object be placed?

What is the diameter of the image?

Part complete

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

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.

Answer 1

The mirror equation is written as follows: lu v Here, u is the object distance and v is the distance of the sun Rewrite the above equation for object distance as follows vf (15x1010 m) (0.47 m) (15x101° m) -(0.47 m) 0.47 m -47 cm Therefore, the object is placed at the focal point of the mirror i.e. 47 cm The expression for magnification is, 0.47 m (14 10 m 10 15x10* m -4.38x10-3 m Therefore, the diameter of the image is -4.38x10 m

The intensity of the incoming sunlight is I-1050 W/m' The expression for power is, The area of the circle in terms of diameter is, -3.14 (0.24 m = 0.045216 m2 The power captured by the mirror is, P= (1050 W/m2)(0.045216 m.) 47.47 W The area of the image is, --(3.14)(-4.38x10-3 m 1.5x10- m2 The expression for the intensity of the sunlight in the projected image is, 47.47 W 15x10. m - 31.6x10 W/m