The sun's diameter is 1.4 times 10^9 m and i | vedclass

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(C) The angular diameter $	heta$ subtended by the sun at the lens is given by $	heta = \frac{	ext{Diameter of sun}}{	ext{Distance of sun}} = \frac

Vedclass · Accepted Answer

$2.8$

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(C) The angular diameter $	heta$ subtended by the sun at the lens is given by $	heta = \frac{	ext{Diameter of sun}}{	ext{Distance of sun}} = \frac{1.4 	imes 10^9 \ m}{10^{11} \ m} = 1.4 	imes 10^{-2} \ 	ext{radians}$.Since the sun is at a very large distance,the image is formed at the focal plane of the convex lens.Let $d$ be the diameter of the image formed at the focal plane. Then,$	heta = \frac{d}{f}$,where $f = 2 \ m$ is the focal length.Therefore,$d = 	heta 	imes f = (1.4 	imes 10^{-2}) 	imes 2 \ m = 2.8 	imes 10^{-2} \ m$.Converting the result into centimeters,$d = 2.8 	imes 10^{-2} 	imes 100 \ cm = 2.8 \ cm$.

The sun's diameter is $1.4 \times 10^9 \ m$ and its distance from the earth is $10^{11} \ m$. The diameter of its image,formed by a convex lens of focal length $2 \ m$,will be.......$cm$.

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