KVPY SX SB Exam 20141 Question Paper

Question : 95 of 120

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A concave mirror of radius of curvature R has a circular outline of radius r. A circular disk 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 < < R, the area of this disc has to be at least

${πr^6 }/{4R^4}$
${πr^4}/{4R^2}$
${πr^5}/{4R^3}$
${πr^4 }/{ R^2}$

Validate

Solution:

d = radius of disc

A = πd^2

From similar triangle

r/d={R-R/{2\cosθ}}/{R/2-({R-R/{2\cosθ}})}

r/d={2\cosθ-1}/{ -\cosθ+ 1}

d=({ -\cosθ+ 1}/{2\cosθ-1}).r

\sin θ = r/R∴ \cosθ= √{R^2-r^2}/R

\cos θ =(1-r^2/R^2)^{1\/2}

\cos θ = 1 -1/2\ r^2/R^2

1 -\cos θ =r^2/{2R^2}

d ={r^2×r}/{2R^2[2(1-r^2/{2R^2})-1]

d = r^3/{2R^2[1-r^2/{2R^2}]}=r^3/{2R^2}

A = πd^ 2 ={πr^6}/{4R^4}

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