$(a)\;u=2R \\ (b)\;u < 3R \\ (c)\;u > 3R \\ (d)\;u < R$

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We know that

$ \large\frac{ \mu_2}{v}=\frac{\mu _1}{u}=\frac{\mu_1-\mu_1}{R}$

Since object is in glass,

$\mu _2=1, \mu_1=1.5$

We have,

$\large\frac{1}{v} -\frac{1.5}{(-u)}=\frac{1-1.5}{-R}$

$\large\frac{1}{v}+\frac{3}{2u} =\frac{1}{2R}$

=> $\large\frac{1}{v} =\frac{1}{2R}-\frac{3}{2u}$

for image to be real,

v is to be positive

$\large\frac{1}{2R} > \frac{3}{2u}$

or $ u > 3R$

Hence c is the correct answer.

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