Question

A diverging meniscus lens of $1.5$ refractive index has concave surfaces 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. 7cm
• B. -8cm
• C. 9cm
• D. 10cm

Answer 1

Answer: (B)

-8cm

Answer 2

$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ For given concave lens $R_{1}=-3\, c m$ and $R_{2}=-4\, c m$ $\therefore \frac{1}{v}-\frac{1}{u}=(\mu-1)\left(\frac{1}{-3}+\frac{1}{4}\right)$ 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 $v=-8\, cm$