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

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(B) Given: Intensity of incident light $I_0 = 1.3 	ext{ kW m}^{-2}$,focal length $f = 20 	ext{ cm}$.Let $R$ be the radius of the lens aperture and $

Vedclass · Accepted Answer

$130$

Vedclass · Answer

(B) Given: Intensity of incident light $I_0 = 1.3 	ext{ kW m}^{-2}$,focal length $f = 20 	ext{ cm}$.Let $R$ be the radius of the lens aperture and $r$ be the radius of the light beam at a distance $d = 22 	ext{ cm}$ from the lens.The distance of this plane from the focus $F$ is $x = d - f = 22 	ext{ cm} - 20 	ext{ cm} = 2 	ext{ cm}$.By similar triangles $\Delta ABF$ and $\Delta PQF$ (where $AB$ is the diameter of the lens and $PQ$ is the diameter of the beam at the given plane):$\frac{r}{R} = \frac{x}{f} = \frac{2 	ext{ cm}}{20 	ext{ cm}} = \frac{1}{10}$.The area of the lens is $A = \pi R^2$ and the area of the beam at the given plane is $a = \pi r^2$.Thus,$\frac{a}{A} = \left(\frac{r}{R}ight)^2 = \left(\frac{1}{10}ight)^2 = \frac{1}{100}$.Since the total power $P$ incident on the lens is conserved and passes through the area $a$,the intensity $I$ at this plane is given by $I 	imes a = I_0 	imes A$.Therefore,$I = I_0 	imes \left(\frac{A}{a}ight) = 1.3 	imes 100 = 130 	ext{ kW m}^{-2}$.

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