Question

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

• A. $7 \,cm$
• B. $-8 \,cm$
• C. $9 \,cm$
• D. $10 \,cm$

Answer 1

Answer: (B)

We have $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
For given concave lens
$R_{1}=-3\, cm$ and $R_{2}=-4\, cm$
$\therefore \frac{1}{v}-\frac{1}{u}=(\mu-1)\left(\frac{1}{-3}+\frac{1}{4}\right)$
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 $\frac{1}{v} =-\frac{1}{24}-\frac{1}{12}$
$=\frac{-1-2}{24}$
$=-\frac{3}{24}=-\frac{1}{8}$
$\Rightarrow V =-8\, cm$