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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Physics
Chapter
Ray Optics

BITSAT 2011 · Physics · Ray Optics

A thin convergent glass lens (μ_ g =1.5) has a power of +5.0 D ....

A thin convergent glass lens (μg=1.5)\left(\mu_{g}=1.5\right) has a power of +5.0D+5.0 \mathrm{D}. When this lens is immersed in a liquid of refractive index μ,\mu, it acts as a divergent lens of focal length 100 cm100 \mathrm{~cm}. The value of μ\mu must be

Options

  1. A4/34 / 3
  2. B5/35 / 3
  3. C5/45 / 4
  4. D6/56 / 5

Correct answer

B. 5/35 / 3

Step-by-step solution

PaP1=(μgμa−1)(μgμ1−1)=+5−100/100=−5−5(μgμ1−1)=μgμa−11.5μ1−1=−15(1.5−1)=−0.1\frac{\mathrm{P}_{\mathrm{a}}}{\mathrm{P}_{1}}=\frac{\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\right)}{\left(\frac{\mu_{\mathrm{g}}}{\mu_{1}}-1\right)}=\frac{+5}{-100 / 100}=-5-5\left(\frac{\mu_{\mathrm{g}}}{\mu_{1}}-1\right)=\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{a}}}-1\frac{1.5}{\mu_{1}}-1=\frac{-1}{5}(1.5-1)=-0.1μ1=1.50.9=53\begin{array}{ll}\mu_{1}=\frac{1.5}{0.9}=\frac{5}{3}\end{array}$$
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