The diameter of the Sun is 1.4 times 10^9 , m | vedclass

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(C) Let $D$ be the diameter of the Sun and $d$ be the diameter of the image formed by the lens.The distance of the Sun from the lens is $u = 10^{11} \

Vedclass · Accepted Answer

$2.8$

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(C) Let $D$ be the diameter of the Sun and $d$ be the diameter of the image formed by the lens.The distance of the Sun from the lens is $u = 10^{11} \, m$.The focal length of the lens is $f = 2 \, m$.Since the Sun is at a very large distance,the image is formed at the focus of the lens.From the geometry of the rays passing through the optical center,we have the relation:$	an \alpha = \frac{D}{u} = \frac{d}{f}$Substituting the given values:$\frac{1.4 	imes 10^9}{10^{11}} = \frac{d}{2}$$d = \frac{2 	imes 1.4 	imes 10^9}{10^{11}} \, m$$d = 2.8 	imes 10^{-2} \, m$Converting to centimeters:$d = 2.8 	imes 10^{-2} 	imes 10^2 \, cm = 2.8 \, cm$.

The diameter of the Sun is $1.4 \times 10^9 \, m$ and its distance from the Earth is $10^{11} \, m$. What is the diameter of the image (in $cm$) formed by a convex lens of focal length $2 \, m$?

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