For convex lenses, let radius of curvature is R. Then,
15cm1=(1.5−1)(R1−−R1)
⇒R1=151⇒R=15cm
Now for concave lens,
fconcave1=(1.25−1)(−R1−R1)
=41×−R2=30−1
feff1=151+151−301=−101
⇒feff=−10cm
Two identical thin biconvex lenses of focal length 15cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is _____ cm.
Held on 24 Jun 2022 · Verified 6 Jul 2026.
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