Question

A spherical convex surface of radius of curvature 2/9 m separate two media of reflective index 4/3 and 3/2 . If an object is kept in first medium at Infinity than the position of image from spherical surfaces is

Answer 1

Step 1: Given

${\eta }_1=\frac{4}{3}$

${\eta }_2=\frac{3}{2}$

Radius of curvature $\left(R\right)=\frac{2}{9}m$

Object is at infinity, so $\left(u\right)=\infty$

Step 2: Formula used and calculation

$-\frac{{\eta }_1}{u}+\frac{{\eta }_2}{v}=\frac{{\eta }_2-{\eta }_1}{R}$

$\frac{\left(\frac{4}{3}\right)}{-\infty }+\frac{\left(\frac{3}{2}\right)}{v}=\frac{\left(\frac{3}{2}\right)-\left(\frac{4}{3}\right)}{\left(\frac{2}{9}\right)}$

$\frac{3}{\left(2v\right)}=\frac{1}{6}\times \frac{9}{2}$

$v=2m$