Question

An illuminated object and a screen are placed $90cm$ apart. The focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object would be:

• A. $10cm$
• B. $20cm$
• C. $15cm$
• D. $30cm$

Answer 1

Answer: (B)

$\therefore u+v=90$ ...(i)
$m=\frac{\upsilon }{u}\Rightarrow 2=\frac{\upsilon }{u}\Rightarrow v=2u$ ...(ii)
Putting the value of v in (i), we get
$u+2u=90\Rightarrow u=\frac{90}{30}=30$
$\therefore v=2\times 30=60$
Using lens formula, we get
$$\begin{gathered} \frac{1}{f}=\frac{1}{\upsilon }-\frac{1}{u} \\ \Rightarrow \frac{1}{f}=\frac{1}{60}-\frac{1}{-30} \\ \Rightarrow \frac{1}{f}=\frac{1}{60}+\frac{1}{30} \\ \Rightarrow \frac{1}{f}=\frac{1+2}{60}=\frac{3}{60} \\ \Rightarrow f=\frac{60}{3}\therefore f=20\text{ cm}\text{.} \end{gathered}$$