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CBSE Class 12 Physics 2023 Outside Delhi Set 3 Paper

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Question : 10 of 13
Marks: +1, -0
The power of a thin lens is +5 D+5\,\mathrm{D}. When it is immersed in a liquid, it behaves like a concave lens of focal length 100 cm100\,\mathrm{cm}. Calculate the refractive index of the liquid. Given refractive index of glass = 1.5
Power of the lens =+5 D=+5\,\mathrm{D}
Focal length of the lens =f=  1D=f=\;\frac{1}{D}
Or,   f=  15 cm\;f=\;\frac{1}{5}\,\mathrm{cm}
∴    f=20 cm\therefore\;\;f=20\,\mathrm{cm}
Using lens makers' formula,
  1fair  =  (μglass−μairμair)  (1R1−1R2)\;\frac{1}{f_{\text{air}}}\;=\;\left(\frac{\mu_{\text{glass}}-\mu_{\text{air}}}{\mu_{\text{air}}}\right)\;\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
Or,   120  =  (1.5−11)  (1R1−1R2)\;\frac{1}{20}\;=\;\left(\frac{1.5-1}{1}\right)\;\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
Or,   120  =0.5  (1R1−1R2)\;\frac{1}{20}\;=0.5\;\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
∴    (1R1−1R2)=  110\therefore\;\; \left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\;\frac{1}{10}
When immersed in liquid,
  1fliquid=(μglass−μliquidμliquid)\;\frac{1}{f_{\text{liquid}}} = \left(\frac{\mu_{\text{glass}}-\mu_{\text{liquid}}}{\mu_{\text{liquid}}}\right) (1R1−1R2)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
  Or,  −  1100=(1.5−μliquidμliquid)\;\text{Or,}\;-\;\frac{1}{100}= \left(\frac{1.5-\mu_{\text{liquid}}}{\mu_{\text{liquid}}}\right) (1R1−1R2)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
  Or,    −  1100=(1.5−μliquidμliquid)\;\text{Or,}\;\; -\;\frac{1}{100}= \left(\frac{1.5-\mu_{\text{liquid}}}{\mu_{\text{liquid}}}\right) (110)\left(\frac{1}{10}\right) [Putting (1R1−1R2)=110\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{1}{10} ]
  Or,    110=(1.5−μliquidμliquid)\;\text{Or,}\;\;\frac{1}{10}= \left(\frac{1.5-\mu_{\text{liquid}}}{\mu_{\text{liquid}}}\right)
  Or,  9μliquid=15\;\text{Or,}\;9\mu_{\text{liquid}}=15
∴  μliquid=  159=  53\therefore\;\mu_{\text{liquid}}=\;\frac{15}{9}=\;\frac{5}{3}
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