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 $$F1$$. When the space between the two lenses is filled with glycerin (which$$ has the same refractive index $$($$μ$$=1.5) as that of $$glass)$$ then the equivalent focal length is $$F2$$. The ratio $$F1$$:$$F2$$ will be

Question

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 $$F1$$. When the space between the two lenses is filled with glycerin (which$$ has the same refractive index $$($$μ$$=1.5) as that of $$glass)$$ then the equivalent focal length is $$F2$$. The ratio $$F1$$:$$F2$$ will be

Infinity Learn · Accepted Answer

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 $$F1F2=12$$

Start JEE / NEET / Foundation preparation at rupees 99/day !!

Detailed Solution

Get Expert Academic Guidance – Connect with a Counselor Today!