A concave mirror of radius of curvature R has a circular outline of radius r. A circular disc is to be placed normal to the axis at the focus so that it collects all the light that is reflected from the mirror from a beam parallel to the axis. For r `gt gt R`, the area of this disc has to be at least

Options

A. `(pi r^(6))/(4 R^(4))`
B. `(pi r^(6))/(4 R^(2))`
C. `(pi r^(5))/(4 R^(3))`
D. `(pi r^(4))/(R^(2))`

Answer

A

Explanation

d `rarr ` Radius of disc
A = `pi d^(2)`
From similar triangle
`(r )/(d) = (R - (R )/(2 cos theta) )/( (R )/(2) - ( R - (R )/(2 cos theta ) ) ) rArr (r )/(d) = (2 cos theta - 1)/( - cos theta + 1)`
`rArr d = ( (-cos theta + 1)/( 2 cos theta + 1)) r `
`sin theta = (r )/(R ) , cos theta = (sqrt(R^(2) - r^(2)))/(R ) `
`cos theta = (1 - (r^(2))/(R^(2)) )^((1)/(2) ) = 1 - (1)/(2) (r^(2))/(R^(2))`
`" "`(Using Binomial expansion )
` 1 - cos theta = (r^(2))/(2 R^(2)) `
`d = (r^(2) xx r )/(2 R^(2) [ 2 ( 1 - (r^(2))/(2R^(2)) ) - 1 ] ) rArr d = (r^(3))/(2 R^(2)) `
Area (A) = `pi d^(2) = (pi r^(6))/(4 R^(4))`.