A thin equiconvex glass lens of refractive index 1.5 has power of 5D.When the lens is immersed in a liquid of refractive index "s" it acts as divergent lens of focal length 10ocm. the value if "s" is??

 According to Lensmaker’s formula, the focal length for a convex lens placed in air can be obtained as

 

1/f = [(n2/n1) – 1] (1/R1 – 1/R2)  where P = 1/f =5 and n2 = 1.5, n­1=1

Therefore  5= (1.5 -1)  (1/R1 – 1/R2) = 0.5(1/R1 – 1/R2)

 

1/f s= [(n2/ns) – 1] (1/R1 – 1/R2)  where P = 1/fs =1/1m =1  and n2 = 1.5, n­1=s

 

Taking ratio of the above two equation and simplifying ,

s = 1.67

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thanks a lot..:)

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Aakash module 7 Pg 94 Q29
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aakash module pg 246 que 29
  • -18
The power of the lens after immersing in the liquid should be -1 as it acts a divergent lens.
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Sorry but equation solve krne par 1.36 aarha hain but tumne iss question ka answer 1.67 btaya h
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Rajendra
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Here is the answer.

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