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NEET PHYSICSMedium

Two similar thin equi-convex lenses, of focal length ff each, are kept coaxially in contact with each other such that the focal length of the combination is F1F_1. When the space between the two lenses is filled with glycerin which has the same refractive index as that of glass (μ=1.5\mu = 1.5), then the equivalent focal length is F2F_2. The ratio F1:F2F_1 : F_2 will be:

A

3 : 4

B

2 : 1

C

1 : 2

D

2 : 3

Step-by-Step Solution

  1. Initial Combination (F1F_1):
  • For an equi-convex lens with refractive index μ=1.5\mu=1.5 and radius of curvature RR, the focal length ff is given by Lens Maker's Formula: 1f=(μ−1)(2R)=(1.5−1)2R=1R\frac{1}{f} = (\mu-1)(\frac{2}{R}) = (1.5-1)\frac{2}{R} = \frac{1}{R}. Thus, f=Rf = R.
  • When two such lenses are in contact, the equivalent focal length F1F_1 is: 1F1=1f+1f=2f\frac{1}{F_1} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}. Therefore, F1=f2F_1 = \frac{f}{2}.
  1. With Glycerin (F2F_2):
  • The space between the two convex lenses forms an equi-concave liquid lens. Since the refractive index of glycerin (1.51.5) is the same as the glass, the liquid lens behaves as a concave lens with the same radius of curvature RR.
  • Focal length of the liquid lens (flf_l): 1fl=(1.5−1)(−1R−1R)=0.5(−2R)=−1R=−1f\frac{1}{f_l} = (1.5-1)(\frac{-1}{R} - \frac{1}{R}) = 0.5(\frac{-2}{R}) = -\frac{1}{R} = -\frac{1}{f}. So, fl=−ff_l = -f.
  • The new system consists of three lenses in contact: Convex (ff) + Concave (fl=−ff_l = -f) + Convex (ff).
  • The equivalent focal length F2F_2 is: 1F2=1f+1−f+1f=1f\frac{1}{F_2} = \frac{1}{f} + \frac{1}{-f} + \frac{1}{f} = \frac{1}{f}. Therefore, F2=fF_2 = f.
  1. Calculate Ratio:
  • Ratio F1:F2=f2:f=1:2F_1 : F_2 = \frac{f}{2} : f = 1 : 2.
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