A convex lens made of glass `(mu_(0) = 3//2)` hs focal length f in air. The image of a object placed infront of it is inverted real and magnified. Now the whole arrangement is immersed in water (`mu_(w) = 4//3)` without changing the distance between object and lens. Then

Options

A. the new focal length will becom 4f
B. the new focal length will become `f//4`
C. new image will be virtual and magnified
D. new image will be real, inverted and smaller in size

Answer

a.,c.

Explanation

`(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.