Correct Answer - a.,c.
`(1)/(f_(air))=((3)/(2)-1)((1)/(R_(1))-(1)/(R_(2)))`
`(1)/(f_(water))=((3//2)/(4//3)-1)((1)/(R_(1))-(1)/(R_(2)))`
From these two equations we get, `f_(water)=4f_(air)=4f`
In air image was inverted, real and magnified. Therefore, object was lying between f and 2f. Now the focal length has changed to 4f. Therefore, the object now lies between pole and focus. Hence, the new image will be virtual and magnified.