$(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.

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