Sunlight of intensity 1.3 ~kW ~m^{-2} is inci | vedclass

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(A) Given: Incident intensity $I_0 = 1.3 ~kW/m^2$,focal length $f = 20 ~cm$.Let $R$ be the radius of the lens aperture and $A = \pi R^2$ be its area.T

Vedclass · Accepted Answer

$130$

Vedclass · Answer

(A) Given: Incident intensity $I_0 = 1.3 ~kW/m^2$,focal length $f = 20 ~cm$.Let $R$ be the radius of the lens aperture and $A = \pi R^2$ be its area.The light rays converge at the focus $F$ and then diverge.At a distance $v = 22 ~cm$ from the lens,the rays form a circular cross-section of radius $r$.Using similar triangles $	riangle ABF$ and $	riangle PQF$,where $AB$ is the diameter of the lens $(2R)$ and $PQ$ is the diameter of the cross-section $(2r)$:$\frac{r}{R} = \frac{v - f}{f} = \frac{22 - 20}{20} = \frac{2}{20} = \frac{1}{10}$.The area of the cross-section at $22 ~cm$ is $a = \pi r^2$.Since $r = R/10$,we have $a = \pi (R/10)^2 = \frac{\pi R^2}{100} = \frac{A}{100}$.The total power $P$ incident on the lens is $P = I_0 	imes A$.This same power passes through the area $a$ at $22 ~cm$.Therefore,the intensity $I$ at $22 ~cm$ is $I = \frac{P}{a} = \frac{I_0 	imes A}{A/100} = 100 	imes I_0$.$I = 100 	imes 1.3 ~kW/m^2 = 130 ~kW/m^2$.

$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y + 5 = 0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x = 1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x + 8y = 27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x = 9$

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