Hello! The answer to this is that the image will disappear . We know that

1/f=(u2/u1 - 1)(1/R1 - 1/R2)

where u1= refractive index of medium of object

u2= refractive index of lens

and R1 and R2 are the radius of curvature of the two surfaces of lens.

We know that:

1/f= 1/v - 1/u and also

1/f= (Ureal - 1)(1/R1 - 1/R2)

where Ureal= refractive index in which the set up is kept.

Now, when we immerse the whole apparatus in water then Ureal changes and as you changes f also changes. Now, the value of u will be same but f changes so the value of v also changes but the SCREEN IS KEPT AT THE SAME DISTANCE AS BEFORE so the image will not coincide on the screen and it will disappear.

Hope the answer helps.