Therefore, μX=43 , μY=32
And, according to sign convention, R=−10cm.
Now, let us consider, the object is situated at a distance of x in front of the concave spherical surface. So, object distance, u=−x (using sign convention)
According to lens make's formula, we know,
μYv−μXu=μY−μXR
Where, v is the image distance.
∴32v−43−x=32−43−10=−160
Or, v=−1.5x0.017x+1.33
So, clearly, v<0
This implies that the image will always be a virtual one.
Therefore, the correct option is - (C) The image is always virtual.