Question

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 $15cm$ 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 $25cm$ from the lens. Calculate the refractive index of the liquid.

• A. $1.6$
• B. $2.6$
• C. $2.8$
• D. $3.2$

Answer 1

The correct option is A $1.6$

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,

$\frac{3}{2v_1}-\frac{1}{-25}=\frac{\frac{3}{2}-1}{R}$

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

object.

$\frac{\mu }{v_2}-\frac{3}{2v_1}=\frac{\mu -\frac{3}{2}}{-R}$

As explained above, $v_2=-\infty$

Thus solving the above equations for $R=10cm$ as found previously gives $\mu =1.6$