Question

Two identical thin biconvex lenses of focal length 15 cm 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.

Answer 1

Correct Answer: 10

\(\frac{1}{f_1}\)=(\(\frac{μ_e}{μ_{m-1}}\))(\(\frac{1}{R_1}-\frac{1}{R_2}\))
here |R₁|=|R₂|=R
⇒\(\frac{1}{fl_1}\)=(1.5-1)(-\(\frac{2}{R}\))=\(\frac{1}{15}\)
⇒\(\frac{1}{R_1}\)=\(\frac{1}{15}\) or R=15 cm
for the concave lens made up of liquid
⇒\(\frac{1}{fl_2}\)=(1.25-1)(-\(\frac{2}{R}\))=-\(\frac{1}{30}\) cm
now for an equivalent lens,
\(\frac{1}{f_e}\)=\(\frac{2}{fl_1}\)+\(\frac{1}{fl_2}\)
=\(\frac{2}{15}\)-\(\frac{1}{30}\)=\(\frac{3}{30}\)=\(\frac{1}{10}\)
or f_e = 10 cm