Question

List - 1 gives different lens configurations. Radius of curvature of each surface is depicted in figure. Refractive index of glass of lens is $\frac{3}{2}$ and medium in which it is placed is 1.2 . Rays of light parallel to axis of lens from left passes through lens and get focused at distance 'x' from lens. List - 2 gives value of 'x' then correct match is.

• A. $I\to Q,II\to Q,III\to T,IV\to T$
• B. $I\to R,II\to R,III\to U,IV\to U$
• C. $I\to Q,II\to Q,III\to S,IV\to S$
• D. $I\to R,II\to R,III\to S,IV\to S$

Answer 1

The correct option is C $I\to Q,II\to Q,III\to S,IV\to S$

For a convex lens, $R_1$ is taken as positive and $R_2$ is taken as negative. For a concave lens, $R_1$ is taken as negative and $R_2$ is taken as positive.
By $\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

I) $\frac{1}{f}=(\frac{1.5}{1.2}-1)(\frac{1}{R}+\frac{2}{R})\Rightarrow f=x=\frac{4R}{3}$

II) $\frac{1}{f}=(\frac{1.5}{1.2}-1)(-\frac{2}{R}-\frac{1}{R})\Rightarrow f=x=-\frac{4R}{3}$

III) $\frac{1}{f}=(\frac{1.5}{1.2}-1)(\frac{1}{\infty }-\frac{1}{R})\Rightarrow f=x=-4R$

IV) $\frac{1}{f}=(\frac{1.5}{1.2}-1)(\frac{2}{R}-\frac{1}{R})\Rightarrow f=x=4R$