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. 7 cm
• B. - 8 cm
• C. 9 cm
• D. 10 cm

Answer 1

Answer: (B)

$\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ For given concave lens, $R_1=-3cm,$ $R_2=-4cm$ $\therefore$ $\frac{1}{v}-\frac{1}{u}=(\mu -1)\left(\frac{1}{-3}+\frac{1}{4}\right)$ 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 $v=-8cm$