Question

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

$\frac{1}{\text{V}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow \frac{1}{\text{v}}-\frac{1}{-15}=\frac{1}{10}\Rightarrow \frac{1}{\text{v}}=\frac{1}{10}-\frac{1}{15}$
$\frac{1}{\text{V}}=\frac{3-2}{30}\Rightarrow \text{v}=30\text{cm}$
$\text{for}\text{r}\text{, }\frac{1\text{.}5}{30}=\frac{\text{r}}{10}\Rightarrow \text{r}=\frac{1}{2}\text{cm}\Rightarrow \text{A}=\pi {\left(\text{r}\right)}^2\Rightarrow \text{A}=\pi {\left(\frac{1}{2}\right)}^2$
$\text{A}=\frac{\pi }{4}{\text{cm}}^2$