Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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 $10 \text{ cm}$ . The diameter of the sun is $1.39 \times 10^9 \text{ m}$ and its mean distance from the earth is $1.5 \times 10^{11} \text{ m}$ . What is the diameter of the sun's image on the paper?

$9.2 \times 10^{-4} \text{ m}$

$6.5 \times 10^{-4} \text{ m}$

$6.5 \times 10^{-5} \text{ m}$

$12.4 \times 10^{-4} \text{ m}$

Step-by-Step Solution

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 $f$ . $v = f = 10 \text{ cm} = 0.1 \text{ m}$
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 ( $u$ ). $m = \frac{h_i}{h_o} = \frac{v}{u}$
Calculation: Substituting the given values:

Object size (Diameter of Sun), $h_o = 1.39 \times 10^9 \text{ m}$
Object distance (Distance of Sun), $u = 1.5 \times 10^{11} \text{ m}$
Image distance, $v = 0.1 \text{ m}$ $h_i = h_o \times \frac{v}{u}$ $h_i = (1.39 \times 10^9) \times \frac{0.1}{1.5 \times 10^{11}}$ $h_i = \frac{1.39 \times 10^8}{1.5 \times 10^{11}} = \frac{1.39}{1.5} \times 10^{-3} \text{ m}$ $h_i \approx 0.9266 \times 10^{-3} \text{ m} = 9.27 \times 10^{-4} \text{ m}$

Conclusion: The calculated diameter is approximately $9.2 \times 10^{-4} \text{ m}$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started