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 mu- it acts as a divergent lens of focal length 100 cm- The value of mu must be4-35-35-46-5

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Toppr Admin · Accepted Answer

PaP1-x3BC-g-x3BC-a-x2212-1-x3BC-g-x3BC-1-x2212-1-5-x2212-100-100-5-x2212-5-x3BC-g-x3BC-1-x2212-1-x3BC-g-x3BC-a-x2212-11-5-x3BC-1-x2212-1-x2212-15-1-5-x2212-1-x2212-0-1-x3BC-1-1-50-9-53

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

$\frac{P_{a}}{P_{1}} = \frac{(\frac{μ_{g}}{μ_{a}} - 1)}{(\frac{μ_{g}}{μ_{1}} - 1)} = \frac{+ 5}{- 100 / 100} = + 5$

$- 5 (\frac{μ_{g}}{μ_{1}} - 1) = \frac{μ_{g}}{μ_{a}} - 1$

$\frac{1.5}{μ_{1}} - 1 = \frac{- 1}{5} (1.5 - 1) = - 0.1; μ_{1} = \frac{1.5}{0.9} = \frac{5}{3}$

A thin convergent glass lens (μg=1.5) has a power of +5.0 D. When this lens is immersed in a liquid of refractive index μ, it acts as a divergent lens of focal length 100 cm. The value of μ must be4/35/35/46/5

PaP1=(μgμa−1)(μgμ1−1)=+5−100/100=+5−5(μgμ1−1)=μgμa−11.5μ1−1=−15(1.5−1)=−0.1;μ1=1.50.9=53

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

$\frac{P_{a}}{P_{1}} = \frac{(\frac{μ_{g}}{μ_{a}} - 1)}{(\frac{μ_{g}}{μ_{1}} - 1)} = \frac{+ 5}{- 100 / 100} = + 5$

$- 5 (\frac{μ_{g}}{μ_{1}} - 1) = \frac{μ_{g}}{μ_{a}} - 1$

$\frac{1.5}{μ_{1}} - 1 = \frac{- 1}{5} (1.5 - 1) = - 0.1; μ_{1} = \frac{1.5}{0.9} = \frac{5}{3}$