Question

A thin equiconvex lens made of glass $(n=\frac{3}{2})$ is placed in such a way, that one surface is in contact with water $(n=\frac{4}{2})$ and another surface is in contact with air. Find focal length of setup if the rays are incident as shown. Radius of curvature $(R=30cm).$

• A. $30cm$
• B. $100cm$
• C. $60cm$
• D. $120cm$

Answer 1

The correct option is C $60cm$


In new convention
$\frac{{\mu }_2}{v}\frac{{\mu }_1}{u}=\frac{{\mu }_2{\mu }_1}{R}$

$\frac{{\mu }_2}{v_1}+\frac{{\mu }_1}{\infty }=\frac{{\mu }_2{\mu }_1}{R}$

$\frac{{\mu }_2}{v_1}=\frac{{\mu }_2-{\mu }_1}{R}$

$v_1=\frac{{\mu }_{2}R}{\left({\mu }_2{\mu }_1\right)}$

The first image will behave like a virtual object for the second surface.
Now the rays are refracted from second surface and will converge at focus
$\frac{{\mu }_3}{f}\frac{{\mu }_2}{v_1}=\frac{{\mu }_3{\mu }_2}{R}$

$\frac{{\mu }_3}{f}\frac{{\mu }_2\left({\mu }_2{\mu }_1\right)}{{\mu }_{2}R}=\frac{{\mu }_3{\mu }_2}{R}$

$\frac{{\mu }_3}{f}=\frac{{\mu }_2{\mu }_1}{R}+\frac{{\mu }_2{\mu }_3}{R}$

$\frac{{\mu }_3}{f}=\frac{({\mu }_2{\mu }_1+{\mu }_2{\mu }_3)}{R}$

$\frac{{\mu }_3}{f}=\frac{\left(2{\mu }_2{\mu }_1{\mu }_3\right)}{R}$

$f=\frac{{\mu }_{3}R}{\left(2{\mu }_2{\mu }_1{\mu }_3\right)}$

Put $f=\frac{{\mu }_{3}R}{\left(2{\mu }_2{\mu }_1{\mu }_3\right)}$, ${\mu }_1=1\text{and}R=30cm,$

$f=60cm$