For a given lens, the magnification was found to be twice as large as when the object was 0.15 m distant from it as when the distance was 0.2 m . The focal length of the lens is

• A

  1.5 m
• Correct answer
  B

  0.20 m
• C

  0.10 m
• D

  0.05 m

Solution

Let as shown, 1 and 2 are positions of objects and images in two different situations. It is given $\left|\frac{v_1}{u_1}\right|=2\left|\frac{v_2}{u_2}\right|$ Here, $u_1=-15 \mathrm{~cm}, u_2=-20 \mathrm{~cm}$ $\therefore \quad v_1=2 v_2 \times \frac{u_1}{u_2}$ $=2 v_2 \times \frac{15}{20}=\frac{3}{2} v_2$ Now, $\quad \frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ $\therefore \quad \frac{1}{f}=\frac{1}{v_1}-\frac{1}{u_1}$ and $\quad \frac{1}{f}=\frac{1}{v_2}-\frac{1}{u_2}$ So, $\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{v_2}-\frac{1}{u_2}$ $\Rightarrow \quad \frac{2}{3 v_2}+\frac{1}{15}=\frac{1}{v_2}+\frac{1}{20}$ $\Rightarrow \quad v=20 \mathrm{~cm}$

Hence, the correct answer is option: 'B'.