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A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the Sun is $1.39 \times 10^{9}$ m and its mean distance from the earth is $1.5 \times 10^{11}$ m. What is the diameter of the Sun's image on the paper ?
A
$6.5 \times 10^{-5}$ m
B
$12.4 \times 10^{-4}$ m
C
$9.2 \times 10^{-4}$ m
D
$6.5 \times 10^{-4}$ m
Detailed Solution
The Sun is very far away, so its rays are practically parallel and the image is formed in the focal plane of the lens.
Object distance $u = 1.5 \times 10^{11}$ m; image distance $v = f = 10$ cm $= 0.1$ m
Magnification: $\dfrac{h_i}{h_o} = \dfrac{v}{u}$
$\Rightarrow \dfrac{h_i}{1.39 \times 10^{9}} = \dfrac{0.1}{1.5 \times 10^{11}}$
$h_i = \dfrac{0.1 \times 1.39 \times 10^{9}}{1.5 \times 10^{11}}$
$h_i = \dfrac{1.39 \times 10^{8}}{1.5 \times 10^{11}}$ m
$h_i = 9.2 \times 10^{-4}$ m
So the diameter of the Sun's image on the paper is about $9.2 \times 10^{-4}$ m.
Object distance $u = 1.5 \times 10^{11}$ m; image distance $v = f = 10$ cm $= 0.1$ m
Magnification: $\dfrac{h_i}{h_o} = \dfrac{v}{u}$
$\Rightarrow \dfrac{h_i}{1.39 \times 10^{9}} = \dfrac{0.1}{1.5 \times 10^{11}}$
$h_i = \dfrac{0.1 \times 1.39 \times 10^{9}}{1.5 \times 10^{11}}$
$h_i = \dfrac{1.39 \times 10^{8}}{1.5 \times 10^{11}}$ m
$h_i = 9.2 \times 10^{-4}$ m
So the diameter of the Sun's image on the paper is about $9.2 \times 10^{-4}$ m.