Questions

# Two similar thin equi-convex lenses, of focal length f each, are kept coaxially in contact with each other such that the focal length of the combination is ${\mathrm{F}}_{1}$. When the space between the two lenses is filled with glycerin (which has the same refractive index $\left(\mathrm{\mu }=1.5\right)$ as that of glass) then the equivalent focal length is ${\mathrm{F}}_{2}$. The ratio ${\mathrm{F}}_{1}$:${\mathrm{F}}_{2}$ will be

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a
3 : 4
b
2 : 1
c
1 : 2
d
2 : 3

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detailed solution

Correct option is C

According to lens maker's formula  1f=(μ-1)1R1-1R2 1f=(μ-1)1R-1-R=(1.5-1)2R=1RTwo similar equi-convex lenses of focal length f each are held in contact with each other.The focal length F1 of the combination is given by1F1=1f+1f=2f; F1=f2=R2                   .........(i)For glycerin in between lenses, there are three lenses, one concave and two convexFocal length of the curve lens is given 1f'=(1.5-1)-2R=-1RNow, equivalent focal length of the combination is given by1F2=1f+1f'+1f;1F2=1R-1R+1R=1RF2=R                                                                      ..........(ii)  Dividing equation (i) by (ii), we get F1F2=12