A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with water of refractive index 4/3. It is found that when a point object is placed 15cm above the lens on its principle axis, the object coincides with its own image. On repeating with another liquid, the object and the image again coincide at a distance 25 cm from the lens. The refractive index of the liquid.

1.6

RI of lens material = $μ_2$; RI of first liquid = $μ_3$; RI of air = $μ_1$; RI of second liquid = $μ'_3$

In general $(μ_3/v) - (μ_1/u) = \{(μ_2 - μ_1)/R\}+ \{(μ_3 - μ_2)/R\}$

Here u = -15 cm, v = ∞

Now, $1/(+15) = [(3/2)-1/R]- [\{(4/3) - (3/2)\}/R]$

⇒ R = 10 cm

For the second liquid, similarly,

$(1/25) = (0.5/R) - [\{μ'_3 - (3/2)\}/R]$

$⇒μ'_3 = 1.6$