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A small point object is inside a glass surface as shown. Refraction takes place at concave spherical boundry separating glass air medium. For the image to be real the object distance u is (R-Radius of curvature of the surface , $\mu of glass =\large\frac{3}{2}$ )


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

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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.
answered Jan 16, 2014 by meena.p

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