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}}=-3\,cm, $ $ {{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{1}{8} $ or $ v=-8\,cm $