The sum subtends an angle `0.5^(@)` at the center of a concave mirror having radius of curvature 1 m. What will be the diameter of the image of the sun formed by the mirror?

Explanation

M M' is a concave mirror [Fig.1.42]. Let the diameter of the sun be D and the distance of the sun from the mirror be u.
`therefore " " (D)/(u)=(pi)/(360)[therefore 0.5^(@)=(1)/(2)xx(pi)/(180)=(pi)/(360)rad]`
`or, " " u=(360D)/(pi)m`
The sun is at a large distance from the mirror. So, the rays coming from the sun are assumed to be parallel and hence its image will be formed at the focal plane.

`therefore " " v=f=(r)/(2)=(1)/(2)=0.5m`
`"Now, magnification", m=-(v)/(u)=-(0.5pi)/(-360D)=-(0.5pi)/(360D)`
[since in the case both u and v are negative]
`"Again", m=(I)/(O)=("diameter of the imager of the sun")/("diameter of the sun"(D))`
`therefore` Diameter of the image of the sun
`= mxxD=(0.5pi)/(360D)xxD=0.004363m=0.4363 cm.`
[taking only the numerical value of m]