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An equi - convex lens of crown and equi convex lens of flint glass make an achromatic system. The radium of curvature of convex lens is $0.54\;m$. If the focal length of combination for mean colour is 1.5 m and $RI$ for crown glass are $\mu _R= 1.53$ and $\mu_v =1.55$ Then the dispersive power of flint glass is

$(a)\;0.4 \\ (b)\;0.55 \\ (c)\;0.65 \\ (d)\;0.7 $

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As for lens
$\large\frac{1}{f}=(\mu-1) \bigg[\large\frac{1}{R_1}-\frac{1}{R_1}\bigg]$
Here $R_1=R_2=R=54\;cm$
and $\mu=\large\frac{1}{2} $$( \mu_v +\mu_R) =1.54$
So for convex lens
$\large\frac{1}{f_i}$$=(1.54-1) \bigg[\large\frac{1}{54}-\frac{1}{-54}\bigg]$
$f_i =50 cm$
Now as for combination of two thin lenses in contact
$\large\frac{1}{f_c} +\frac{1}{f_F}=\frac{1}{F}$
So, $\large\frac{1}{50}+\frac{1}{f_F}=\frac{1}{150}$
$\large\frac{1}{f_F}=\frac{1}{150}-\frac{1}{50}$
$f_F=-75\;cm$
As system is achromatic
$\large\frac{w_c}{f_c}+\frac{w_f}{w_F} $$=0$
ie $w_F=-W_c \large\frac{f_E}{f_c}$
$w_c= \large\frac{\mu_v -\mu R}{\mu-1}$
$\qquad= \large\frac{1.33-1.53}{1.34-1}$
$\qquad= \large\frac{1}{27}$
So substituting values we get,
$w_F=\large\frac{-1}{27} \times \bigg[ \large\frac{-75}{+50}\bigg] $
$\qquad= \large\frac{1}{18}$
$\qquad= 0.55$
Hence b is the correct answer.
answered Feb 14, 2014 by meena.p
 

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