$(a)\;18.14\;cm \\ (b)\;17.14\;cm \\ (c)\;16.14\;cm \\ (d)\;15.14\;cm $

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First surface $\mu_2=1.5, \mu_1=1, \mu=-\infty, \rho=+10\;cm$

$\therefore \large\frac{1.5}{v} -\frac{1}{-\infty}=\frac{1}{20} =\frac{1.5 -1}{10}$

or $\large\frac{1.5}{v}=\frac{1}{20} => $$v=30\;cm$

Second surface,

$\mu_2=-1,\mu_1=1.5, u=30.5=25\;cm$

$R= 5 \;cm$

$\therefore \large\frac{1}{v}-\frac{1.5}{2.5}=\frac{1-1.5}{5}$

or $\large\frac{1}{v}-\frac{3}{50}=\frac{-1}{10}$$=>v= -25\;cm$

Third surface,

$\mu_3=1.5,\mu_1=1, u=-25-10=-35\;cm$

$R=- 5 \;cm$

$\therefore \large\frac{1.5}{v}-\frac{1}{-35}=\frac{0.15-1}{-5}$

or $\large\frac{1.5}{v}+\frac{1}{35}=\frac{-1}{10}$$=>v= -\large\frac{35}{3}$

Fourth surface,

$u= -\large\frac{35}{3}-5 =\large\frac{-50}{3}$

$\mu_2=1,\mu_1=1.5, u= 9-4=5\;cm$

$R= -10\;cm$

$\therefore \large\frac{1}{v}+\frac{(1.5)3}{50}=\frac{1-1.5}{-10}$

$\large\frac{1}{v}-\frac{9}{100}=\frac{1}{20}$

$v=7.14\;cm$

The final position is $10+7.14=17.14\;cm$ (towards right)

Hence b is the correct answer

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