When viewed from the water, the friend will seem taller than his usual height.Let actual height be h and the apparent height be h'.Here, the refraction is taking place from rarer to denser medium and a virtual image is formed.Using −uμ1​​+vμ2​​=Rμ2​−μ1​​Where refractive index of water is μ2​  and refractive index of air is μ1​.u and v are object and image distances, respectively.R  is the radius of curvature, here we will take it as ∞.​−uμ1​​+vμ2​​=∞μ2​−μ1​​uμ1​​=vμ2​​v=μ1​μ2​​×u​ We know magnification is given by:m=uv​Putting the value of v in the above equation:​m=uu×μ2​​m=μ2​​ As the magnification is greater than 1 , so the apparent height seems to be greater than actual height.