EE1002: Derive a Lens Maker’s Equation to Express the Lens Behavior of the Air: Physics Assignment, NTU, Singapore

Assignment Overview:

1. Image formation by refraction is governed by:

where the symbols deliver the usual meaning as in the lecture notes. Consider an air pocket immersed in water as shown in Figure 1. The air pocket works as a diverging lens in such a condition.

(i) Derive a lens maker’s equation to express the lens behavior of the air pocket, starting from the equation given above. Your working should include every step, with the aid of a diagram whenever necessary, to arrive at the lens maker’s equation. Assume that the refractive index of air is 1.00.

(ii) Determine the focal length of the air pocket when both the curvatures have the same radius of 5 mm. Assume that the water has a refractive index of 1.33.

2. A parallel-plate capacitor is formed by two identical square metal plates separated by a distance d = 0.4 cm, as shown in Figure 2. Each side of the metal plate has a length l = 10 cm. A dielectric slab with the same cross-sectional area as the metal plates is placed inside the capacitor. The slab has a dielectric constant k= 4 and a thickness of 0.2 cm. The capacitor is fully charged by a 120 V battery. At t = 0, the dielectric slab is located completely within the capacitor (i.e. x = l). At t >= 0, an external force F is applied, pulling the slab out of the capacitor at a constant speed v0 = 2 cm/s. During this process, the distances from the two metal plates to the top and bottom surfaces of the slab are fixed as 0.1 cm. The capacitor is connected to the battery at all times.

(i) Find the capacitance and the charge on the capacitor at t = 0.

(ii) Determine the time to when the right edge of the slab is just being pulled out of the capacitor, that is, to is the time when the slab is just totally out of the capacitor. Express the capacitance and the charge on the capacitor as a function of t for 0 ≤ t ≤ to and t ≥ to.

(iii) Comment on how the charge and voltage of the capacitor change with time when the slab is being pulled out.

3. As shown in Figure 3, a uniform magnetic field B = 0.4 T exists in the region x > 0. The magnetic field is perpendicular to the x-y plane and points into the page. A rectangular superconducting loop (the resistance of which is zero) enters the magnetic field with an initial velocity v0 parallel to the +x axis. At a given time t, the loop has travelled a distance d into the magnetic field and that d = 0 at t = 0. The superconducting loop has a mass m = 30 g and an inductance L = 1.2 mH. The edges l1 = 0.2 m and l2 = 0.3 m are parallel to the x and y axes, respectively. Assume that the loop only experiences the magnetic force and neglect all other external forces, such as friction.

(i) Determine the direction of the net magnetic force acting on the loop when it enters the magnetic field.

(ii) Find the current induced in the loop as a function of the distance d. (Hint: In a closed circuit, the motional emf equals the self-induced emf in the coil)

(iii) Applying the principle of conservation of energy, find the velocity of the loop as a function of the distance d. Hence determine the minimum initial velocity v0 required for the entire loop to enter into the magnetic field.

(iv) If v0 = 2.5 m/s, the loop will stop at a certain d before it completely enters the magnetic field. Find the distance d when the loop stops completely. Comment on whether the loop can remain stationary at this position.

(v) If v0 = 5 m/s, find the terminal velocity of the loop and the corresponding magnetic energy stored in it when the loop has completely entered into the magnetic field.