A thin plano convex lens made of glass of refractive index 1.5 is immersed in a liquid of refractive index 1.2. When the plane side of the lens is silver coated for complete reflection, the lens immersed in the liquid behaves like a concave mirror of focal length 0.2 m. The radius of curvature of the curved surface of the lens is :-

Step 1: Effective Focal Length of a Silvered Plano-Convex Lens

When a thin plano-convex lens is silvered on its plane side, it behaves as a mirror-lens combination with an effective focal length given by:

$$\frac{1}{F} = \frac{2}{f_L} + \frac{1}{f_M}$$

where:

• $F$ is the effective focal length of the system (given as $-0.2$ m),
• $f_L$ is the focal length of the lens in the liquid,
• $f_M$ is the focal length of the mirror (which is at the plane surface, so $f_M = \infty$).

Since $f_M = \infty$, the equation simplifies to:

$$\frac{1}{F} = \frac{2}{f_L}$$

 $$f_L = 2F$$ $$f_L = 2 \times (-0.2) = -0.4 \text{ m}$$

(The negative sign indicates a diverging lens in the liquid.)

Step 2: Lensmaker’s Formula in a Liquid

The focal length of a lens in a medium of refractive index $n_l$ is given by:

$$\frac{1}{f_L} = \left(\frac{n_g}{n_l} - 1\right) \left(\frac{1}{R} - 0\right)$$

Since the second surface is plane, its radius of curvature is infinite, so its contribution is zero.

$$\frac{1}{-0.4} = \left(\frac{1.5}{1.2} - 1\right) \frac{1}{R}$$

 $$-2.5 = \left(\frac{1.5}{1.2} - 1\right) \frac{1}{R}$$ $$-2.5 = \left(\frac{1.5 - 1.2}{1.2}\right) \frac{1}{R}$$ $$-2.5 = \left(\frac{0.3}{1.2}\right) \frac{1}{R}$$ $$-2.5 = \frac{0.3}{1.2R}$$ $$R = \frac{0.3}{2.5 \times 1.2}$$ $$R = \frac{0.3}{3}$$ = 0.1 m 

Final Answer:

The radius of curvature of the curved surface of the lens is 0.1 m