If a lens of focal length f is divided into two equal parts as shown in figure (1) and each has a focal length f′ then ‌
1
f
=‌
1
f′
+‌
1
f′
‌ i.e., ‌f′=2f i.e., each part will have focal length 2f. Now if these parts are put in contact as in figure (2), then resultant focal length of the combination will be ' ‌
1
F
=‌
1
2f
+‌
1
2f
‌ i.e., ‌F=f‌‌‌ (initial value) ‌
For this combination, ‌
1
F
=(aµg−1)(‌
1
R1
−‌
1
R2
) . . . (i) Now, if this combination is immersed in liquid, then ‌