Question

Explain what happens when a convex lens of refractive index 1.2 is immersed in a liquid of refractive index 1.3.

Answer 1

when a convex lens of refractive index u immersed in medium of refractive index Um then we have following cases :

Case i :
======

(1) u = Um => lense behave like a plane sheet of glass Reason :
According to lens makers formula :
1 / f(air ) = ( ug - 1 ) ( 1 / R1 - 1 / R2 )
Where ug = refractive index of glass w.r.t air
R1 and R2 are the Radii of curvature of the two covex surfaces..

When medium is introduced to lens then we have
1 / fm = ( ug / um - 1 ) ( 1 / R1 - 1 / R2 ) --- ( i )
Where fm = focal length of the lens in medium ( water , kerosene etc..)
Note : R1 and R2 will not changed at all when a lens is being immersed in a medium...
um = refractive index of a medium

Since ug = um then equation ( i ) changes to
1 / fm = ( um - um ) ( 1 / R1 - 1 / R2 )
= 0 × ( 1 / R1 - 1 / R2 ) = 0
=> 1 / fm = 0
=> fm = 1 / 0 = infinity
Here focal length is infinity this happens only rays are parallel and so lens behave like a plane glass sheet...

Case ii :
=======
When Ug > um
=> ug > um
=> ug / um > 1
=> ug / um - 1 > 0
=> ug / um is positive
Also ( 1 / R1 - 1 / R2 ) is positive Thus
( ug / um - 1 ) ( 1 / R1 - 1 / R2 ) is positive
=> 1 / fm is positive
=> fm is positive
=> convex lens have positive focal length
=> so lense nature remain same

Case 3 :
=======
When Ug < um
=> ug < um
=> ug / um < 1
=> ug / um - 1 < 0
=> ug / um is negative
Also ( 1 / R1 - 1 / R2 ) is positive Thus
( ug / um - 1 ) ( 1 / R1 - 1 / R2 ) is negative
=> 1 / fm is negative
=> fm is negative
=> concave lens have negative focal length
=> Convex lens behave like a concave lense or show diverging nature...

According to problem
u ( material filled in between the lens )= 1.2 and um = 1.3
Here u < um => as per case ii mentioned above, the the nature of convex lens gets reversed i .e , lens behave like a concave lens...

Answer 2

Answer:

Explanation:

Refer the below attachment.