A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of a focal l

Question

A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of a focal length of $10 	ext{ cm}$. The diameter of the sun is $1.39 	imes 10^9 	ext{ m}$ and its mean distance from the earth is $1.5 	imes 10^{11} 	ext{ m}$. What is the diameter of the sun's image on the paper?

Accepted Answer

1. **Image Formation:** Since the sun is at a very large distance ($u \approx \infty$) compared to the focal length of the lens, the rays from the sun are effectively parallel. These rays converge at the focal plane of the lens. Thus, the image distance $v$ is equal to the focal length $f$.
   $$ v = f = 10 	ext{ cm} = 0.1 	ext{ m} $$
2. **Magnification Formula:** The magnification $m$ produced by a lens is the ratio of the image size ($h_i$) to the object size ($h_o$), which is also equal to the ratio of the image distance ($v$) to the object distance ($u$).
   $$ m = \frac{h_i}{h_o} = \frac{v}{u} $$
3. **Calculation:** Substituting the given values:
   - Object size (Diameter of Sun), $h_o = 1.39 	imes 10^9 	ext{ m}$
   - Object distance (Distance of Sun), $u = 1.5 	imes 10^{11} 	ext{ m}$
   - Image distance, $v = 0.1 	ext{ m}$
   $$ h_i = h_o 	imes \frac{v}{u} $$
   $$ h_i = (1.39 	imes 10^9) 	imes \frac{0.1}{1.5 	imes 10^{11}} $$
   $$ h_i = \frac{1.39 	imes 10^8}{1.5 	imes 10^{11}} = \frac{1.39}{1.5} 	imes 10^{-3} 	ext{ m} $$
   $$ h_i \approx 0.9266 	imes 10^{-3} 	ext{ m} = 9.27 	imes 10^{-4} 	ext{ m} $$
4. **Conclusion:** The calculated diameter is approximately $9.2 	imes 10^{-4} 	ext{ m}$.

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