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InterviewSolution
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A lens has a power of `+-` diopeters in air. What will be its power if completely immersed in water? `(._amu_(w)=4//3 and _amu_(g)=3//2)` |
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Answer» Let `f_(a)` and `f_(w)` be the focal lengths of the lens in air and water, respectively, then `P_(a)=(1)/(f_(a)) or +-5 =(1)/(f_(a))` `f_(a)=0.2 m =20cm` Now, `(1)/(f_(a))=(._amu_(g)-1)[(1)/(R_(1))-(1)/(R_(2))]` (i) and `(1)/(f_(w))=(._wmu_g-1)[(1)/(R_(1))-(1)/(R_(2))]` (ii) Dividing Eq. (i) by Eq. (ii), we get `(f_(w))/(f_(a))=[(._amu_(g)-1)/(._wmu_(g)-1)]` Again, `._wmu _(g)=(._amu_(g))/(._amu_(w))=(3//2)/(1//8)=(9)/(8)` `rArr(f_(w))/(f_(a))=((3//2)-1)/((9//8)-1)=((1//2))/((1//8))=4` `f_(w)=f_(a)xx5=20xx4=80cm=0.8m` `P_(w)=(1)/(f_(w))=(1)/(0.8)=1.25 diopter. ` |
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