A convex lens made of glass `(mu_(g)=3//2)` has focal length f in air. The image of an object placed in front of it is inverted, real and magnified. Now the whole arrangement 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.