A concave mirror has radius of curvature of \(40\, cm\). It is at the bottom of a glass that has water filled up to \(5\, cm\) (see figure). If a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance \(d\) from the surface of water. The value of \(d\) is close to ......\(cm\) (Refractive index of water \(= 1.33\))

\({\text{u}} =  - 5\,{\text{cm}},\) \({\text{f}} = {\text{m}} - \frac{{\text{R}}}{2}\) \( =  - 20\,{\text{cm}}\)

\(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) ; \(\boxed{{v_1} =  + \frac{{20}}{3}\,\,cm}\)

This image will act as object for light getting refracted at water surface.

So, object distance \(d=5+\frac{20}{3}=\frac{35}{3} \,\mathrm{cm}\)

Below water surface.

After refraction, final image is at 

\({d^\prime } = d\left( {\frac{{{\mu _2}}}{{{\mu _1}}}} \right)\) \( = \left( {\frac{{35}}{3}} \right)\left( {\frac{1}{{4/3}}} \right)\)

\(=\frac{35}{4}=8.75 \,\mathrm{cm}\)

\( \approx 8.8\, \mathrm{cm}\)