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(A) For a large aperture mirror,rays parallel to the axis do not converge at a single point (focus $F$) due to spherical aberration. Instead,they form

Vedclass · Accepted Answer

$\frac{\pi r^6}{4 R^4}$

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(A) For a large aperture mirror,rays parallel to the axis do not converge at a single point (focus $F$) due to spherical aberration. Instead,they form a circular patch at the focal plane.Let $N$ be a point on the periphery of the mirror at a distance $r$ from the axis. Let the ray reflected from $N$ intersect the axis at $Q$. In $	riangle NQC$,where $C$ is the center of curvature,the angle of incidence equals the angle of reflection,both being $	heta$. Thus,$\angle N C P = 	heta$ and $\angle N Q C = 	heta$.In $	riangle NQC$,by the law of sines or geometry,$QC = \frac{R}{2 \cos 	heta}$.The distance of $Q$ from the pole $P$ is $PQ = R - QC = R - \frac{R}{2 \cos 	heta}$.The distance of $Q$ from the focus $F$ is $QF = PF - PQ = \frac{R}{2} - (R - \frac{R}{2 \cos 	heta}) = \frac{R}{2 \cos 	heta} - \frac{R}{2} = \frac{R}{2} (\sec 	heta - 1)$.Let $d$ be the radius of the disc at the focus. From similar triangles formed by the reflected ray,the radius $d$ is given by $d = QF 	an(2	heta)$.For small $	heta$,$\sin 	heta \approx 	an 	heta \approx \frac{r}{R}$. Also,$\sec 	heta = (1 - \sin^2 	heta)^{-1/2} \approx 1 + \frac{	heta^2}{2} = 1 + \frac{r^2}{2R^2}$.Substituting these,$QF \approx \frac{R}{2} (1 + \frac{r^2}{2R^2} - 1) = \frac{r^2}{4R}$.Since $	an(2	heta) \approx 2	heta \approx 2(\frac{r}{R})$,we have $d = QF \cdot 2	heta = (\frac{r^2}{4R}) \cdot (\frac{2r}{R}) = \frac{r^3}{2R^2}$.The area of the disc is $A = \pi d^2 = \pi (\frac{r^3}{2R^2})^2 = \frac{\pi r^6}{4R^4}$.

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