Question

A thin equiconvex lens is placed on a horizontal plane mirror and a pin held 20 cm above the lens. The system forms an image that coincides in position with the position of the pin. Now the space between the lens and mirror is filled will water \((μ=\frac{4}{3})\) and then to coincide the pin with will its own image, the pin has to be raised until its distance from the lens is 30 cm. What is the radius of curvature of lens?

Answer 1

The image of pin coincides with of pin if incident rays from pin after refraction through lens fall normally on plane mirror as shown in Fig. Pin is position of focus of the lens.

∴ f₁ = 20 cm

Let the focal length of water lens be f₂ , then for the combination of the two lens, the equivalent focal length F of the combination is 30 cm. We have

where μ is refractive index of water. For plano-convex water lens R₁ = - c, R2 = ∞; therefore