Question

An illuminated object and a screen are placed $90 \, cm$ 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)
$m = \frac{\upsilon}{u} \Rightarrow 2 = \frac{\upsilon}{u} \Rightarrow v = 2 u$ ...(ii)
Putting the value of v in (i), we get
$u + 2 u = 90 \Rightarrow u = \frac{90}{30} = 30$
$\therefore v = 2 \times 30 = 60$
Using lens formula, we get
$ \, \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{.}$