Question

A diverging meniscus lens of $1.5$ refractive index has concave surfaces of radii $3$ and $4cm.$ The position of the image, if an object is placed $12cm$ in front of the lens, is

• A. 7cm
• B. -8cm
• C. 9cm
• D. 10cm

Answer 1

Answer: (B)

$\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
For given concave lens
$R_1=-3cm$ and $R_2=-4cm$
$\therefore \frac{1}{v}-\frac{1}{u}=(\mu -1)\left(\frac{1}{-3}+\frac{1}{4}\right)$
or $\frac{1}{v}-\frac{1}{(-12)}=(1.5-1)\left(\frac{-4+3}{12}\right)$
or $\frac{1}{v}+\frac{1}{12}=0.5\times \frac{-1}{12}=\frac{-1}{24}$
or $\frac{1}{v}=-\frac{1}{24}-\frac{1}{12}=\frac{-1-2}{24}=-\frac{1}{8}$
or $v=-8cm$