Question

A convex lens forms inverted image of a real object on a fixed screen. The size of image is $9 \, cm$ . When lens is displaced $40 \, cm$ along principle axis it again forms a real image of size $4 \, cm$ on the screen. Focal length of the lens is .

• A. 48 cm
• B. 100 cm
• C. 30 cm
• D. 10 cm

Answer 1

Answer: (A)

Since $ \, \, m_{1} \, m_{2}=1 \, \, \, \Rightarrow \, \, \, \left(\frac{h_{1}}{h_{0}}\right) \, \left(\frac{h_{2}}{h_{0}}\right)=1$
$h_{0}= \, \sqrt{h_{1} \, h_{2}}=6cm$
for first position of the lens
$m=-\frac{3}{2}=- \, \frac{x \, + \, 40}{x}$
$\Rightarrow \, \, \, 3x=2x+80$
$\Rightarrow \, \, \, x= \, 80$
$u=-x$ , $y=\left(\right.40+x\left.\right)$
$\frac{1}{f}=\frac{1}{120}-\frac{1}{- 80}$
$\Rightarrow f=48cm$