A thin concavo-convex glass lens (refractive index 1.5 ) features a...

For the glass lens, the top surface is concave and the bottom surface is convex. Since it rests on the convex surface, light from the pin enters the concave surface. Taking the downward direction as positive, the radius of the first surface is $R_1 = -60$ cm and the radius of the second surface is $R_2 = -20$ cm.

The power of the glass lens is given by:
$P_g = (\mu_g - 1)$\left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)<br>P_g = (1.5 - 1) \left( \dfrac{1}{-60} - \dfrac{1}{-20} \right) =0.5 \left( \dfrac{1}{20} - \dfrac{1}{60} \right) = \dfrac{1}{60} \text{ cm}^{-1}<br><br>The bottom surface is silvered and acts as a mirror for light traveling inside the glass. Its radius of curvature isR_2 = -20$cm.<br>Power of the equivalent mirror:<br>$P_m = $-\dfrac{2}{R_2} = -\dfrac{2}{-20} = \dfrac{1}{10} \text{ cm}^{-1}<br><br>The equivalent power of the initial system is:<br>$P = 2P_g + P_m = 2 \left( \dfrac{1}{60} \right) + \dfrac{1}{10} = \dfrac{1}{30} + \dfrac{1}{10} = \dfrac{4}{30} = \dfrac{2}{15} \text{ cm}^{-1}<br><br>The equivalent focal length isF = $\dfrac{15}{2} =$7.5$cm. For the image to coincide with the pin, the pin must be at the center of curvature of the equivalent mirror.<br>Initial position$u_1 = 2F = 15$cm.<br><br>When water is filled in the concave section, a plano-convex water lens is formed with$R_{w1} = \infty and $R_{w2} = -60$ cm.
Power of the water lens:
P_w = (\mu_w - 1) \left(\dfrac{1}{R_{w1}} - \dfrac{1}{R_{w2}} \right) = \left( \dfrac{4}{3} -1 \right) \left( 0 -\dfrac{1}{-60} \right) = \dfrac{1}{180} \text{ cm}^{-1}

The new equivalent power of the system is:
P' = 2P_w + P = 2 \left(\dfrac{1}{180} \right) + \dfrac{2}{15} = \dfrac{1}{90} + \dfrac{12}{90} = \dfrac{13}{90} \text{ cm}^{-1}

The new equivalent focal length is $F' =$\dfrac{90}{13} cm.
New position for the image to coincide is $u_2 = 2F' =$\dfrac{180}{13} \approx$13.85$ cm.

Distance the pin must be moved is:
$\Delta u = u_1 - u_2 = 15 -$\dfrac{180}{13} = \dfrac{15}{13} \approx$1.15 \text{ cm}$

Since $u_2 < u_1$, the pin must be moved towards the lens. The closest matching option provided is $1.14 \text{ cm}$ towards the lens.