Observe the setup illustrated in the figure. The base of the pot...

The eye observes two images of the fish: one formed by direct light rays and the other formed by rays reflected from the mirror at the base.

For the direct image, the real depth of the fish from the water surface is $H - \dfrac{H}{2} = \dfrac{H}{2}$.

The apparent depth of this image from the water surface is $d_1 = \dfrac{H}{2\mu}$.

The distance of this first image from the eye is $D_1 = H + d_1 = H + \dfrac{H}{2\mu} = H\left(1 + \dfrac{1}{2\mu}\right)$ below the eye.

For the reflected image, the fish is at a distance of $\dfrac{H}{2}$ above the plane mirror at the base.

The mirror forms a virtual image of the fish at a distance of $\dfrac{H}{2}$ below the base.

The real depth of this reflected image from the water surface is $H + \dfrac{H}{2} = \dfrac{3H}{2}$.

The apparent depth of this reflected image from the water surface is $d_2 = \dfrac{3H}{2\mu}$.

The distance of this second image from the eye is $D_2 = H + d_2 = H + \dfrac{3H}{2\mu} = H\left(1 + \dfrac{3}{2\mu}\right)$ below the eye.