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A concave lens of glass, refractive index 1.5, has both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a |
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Answer» convergent LENS of focal length 3.5R `(1)/(F)=(._(G)^(m)mu-1)((1)/(R_(1))-(1)/(R_(2)))` Now, `._(g)^(m)mu=(._(g)mu)/(._(m)mu)=(1.5)/(1.75)` For concave lens, as shown in figure, in this case `R_(1)=-R` and `R_(2)=+R` `:.(1)/(f)=((1.5)/(1.75)-1)(-(1)/(R)-(1)/(R))=+(0.25xx2)/(1.75R)` `RARR f=+3.5R` Thepositive sign shows that the lens behaves as a convergent lens. (a) is the correct OPTION. 
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