Question

A thin convergent glass lens ($\mu_{g}$ = 1.5) has a power of + 5.0 D. When this lens is immersed in a liquid of refractive index m, it acts as a divergent lens of focal length 100 cm. The value of m must be(a) 4/3(b) 5/3(c) 5/4(d) 6/5

Answer 1

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(d) 6/5

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Answer 2

Answer:

Explanation:

Power (1/f)= 5 D μ₂ = 1.5 , μ₁ = 1

Focal length -

1/f = (μ₂/μ₁ - 1)[1/R₁ - 1/R₂]

5 = (1.5/1 - 1)[1/R₁ - 1/R₂ ]

5 = (0.5)[ 1/R₁ - 1/R₂ ] eq 1

f = 100 cm = 1m, μ₂ = 1.5 , Let μ₁ = x

1/1 = (1.5/x - 1)[ 1/R₁ - 1/R₂ ]

1 = (1.5/x - 1)[ 1/R₁ - 1/R₂ ] eq 2

Dividing equations (1) and (2),

5 = (0.5)/(1.5/x - 1)

7.5/x - 5 = 0.5

7.5/x = 5.5 ⇒x = 7.5/5.5 = 1.37

Thus, the refractive index of liquid is 1.37