**Given: **

Radius of curvature, R=-5 cm

Object distance, u=-3 cm

The refractive indices n_{1}=1.5 (glass), n_{2}=1.0(air)

We have, \(\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\) (The Lensmaker's formula)

\(\frac{1}{v}-\frac{1.5}{-3}=\frac{1-1.5}{-5}\)

\(\frac{1}{v}-\frac{0.5}{5}=\frac{1.5}{3}=-\frac{2}{5}\)

v = -2.5 cm

Then the image will appear at distance of 2.5 cm from the plane surface of the mirror

**Hence D is the correct option.**