A thin equiconvex lens of refractive index $3 / 2$ is placed on a horizontal plane mirror as shown in figure. The space between the lens and the mirror is filled with a liquid of refractive index $4 / 3$ . It is found what when a point object is placed $15 c m$ above the lens on its principal axis, the object coincides with its own image.
If another liquid is filled instead of water, the object and the image coincide at a distance $25 c m$ from the lens. Calculate the refractive index of the liquid.

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Answer 1

Since the final image coincides with the object the image incident on the mirror must be at its center of curvature, that is, at infinity.

Consider refraction from first surface,

$\frac{3}{2 v _{1}} - \frac{1}{- 2 5} = \frac{\frac{3}{2} - 1}{R}$

Consider refraction from second surface, where image of first acts as its
object.

$\frac{μ}{v _{2}} - \frac{3}{2 v _{1}} = \frac{μ - \frac{3}{2}}{- R}$

As explained above, $v_{2} = - \infty$

Thus solving the above equations for $R = 10 c m$ as found previously gives $μ = 1.6$

Answer 2

Since the final image coincides with the object the image incident on the mirror must be at its center of curvature, that is, at infinity.

Consider refraction from first surface,

\frac{3}{2 v _{1}} - \frac{1}{- 2 5} = \frac{\frac{3}{2} - 1}{R}

Consider refraction from second surface, where image of first acts as its

object.

\frac{μ}{v _{2}} - \frac{3}{2 v _{1}} = \frac{μ - \frac{3}{2}}{- R}

As explained above,

v_{2} = - \infty

Thus solving the above equations for

R = 10 c m

as found previously gives

μ = 1.6