Question

A point source of light is kept at a distance of $15 \, cm$ from a converging lens, on its optical axis. The focal length of the lens is $10 \, cm$ and its diameter is $3 \, cm$ . A screen is placed on the other side of the lens, perpendicular to the axis of the lens, at a distance $20 \, cm$ . from it. Then find the area of the illuminated part of the screen?

• A. $\frac{\pi }{4}\text{cm}^{2}$
• B. $\frac{\pi }{6}\text{cm}^{2}$
• C. $\frac{\pi }{2} \, \text{cm}^{2}$
• D. $\frac{\pi }{3} \, \text{cm}^{2}$

Answer 1

Answer: (A)

$\frac{1}{\text{V}} - \frac{1}{\text{u}} = \frac{1}{\text{f}} \Rightarrow \frac{1}{\text{v}} - \frac{1}{- 1 5} = \frac{1}{1 0} \Rightarrow \frac{1}{\text{v}} = \frac{1}{1 0} - \frac{1}{1 5}$
$\frac{1}{\text{V}}=\frac{3 - 2}{3 0}\Rightarrow \text{v}=30\text{cm}$
$\text{for} \text{r} \text{, } \frac{1 \text{.} 5}{3 0} = \frac{\text{r}}{1 0} \Rightarrow \text{r} = \frac{1}{2} \text{cm} \Rightarrow \text{A} = \pi \left(\text{r}\right)^{2} \Rightarrow \text{A} = \pi \left(\frac{1}{2}\right)^{2}$
$\text{A} = \frac{\pi }{4} \text{cm}^{2}$