Question

Four combinations of two thin lenses are given in Column-I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in Column-II with their focal length in Column–II and select the correct answer using the code given below the column.

• A. (a) – (p); (b) – (q); (c) – (r); (d) – (s)
• B. (a) – (q); (b) – (s); (c) – (r); (d) – (p)
• C. (a) – (s); (b) – (p); (c) – (q); (d) – (r)
• D. (a) – (q); (b) – (p); (c) – (r); (d) – (s)

Answer 1

The correct option is B (a) – (q); (b) – (s); (c) – (r); (d) – (p)


$\frac{1}{f}=(\frac{3}{2}-1)(\frac{1}{r}+\frac{1}{r})=\frac{1}{r}$

$\Rightarrow f=r$

$\frac{1}{f_{eq}}=\frac{1}{f}+\frac{1}{f}=\frac{1}{f}=\frac{2}{r}$

$\Rightarrow f_{eq}=\frac{r}{2}\therefore P\to 2$

$\frac{1}{f}=(\frac{3}{2}-1)\left(\frac{1}{\infty +\frac{1}{r}}\right)=\frac{1}{2r}$

$\Rightarrow f=2r$

$\frac{1}{f_{eq}}=\frac{1}{f}+\frac{1}{f}=\frac{2}{f}=\frac{1}{r}$

$\Rightarrow f_{eq}=r\therefore Q\Rightarrow 4$

$\frac{1}{f_{eq}}=(\frac{3}{2}-1)(\frac{1}{\infty }-\frac{1}{r})=\frac{1}{2r}$

$\Rightarrow f=-2r$

$\frac{1}{f_{eq}}==\frac{1}{f}+\frac{1}{f}=-\frac{2}{2r}$

$\Rightarrow f_{eq}=-r\therefore R\to 3$


$\frac{1}{f_{eq}}==\frac{1}{r}+\frac{1}{-2r}=\frac{1}{2r}$

$\Rightarrow f_{eq}=2r\therefore S\to 1$