Antireflection coatings can be used on the inner surfaces of eyeglasses to reduce the reflection of stray light into the eye, thus reducing eyestrain. You may want to review (Page) . Part A A 90-nm-thick coating is applied to the lens. What must be the coating's index of refraction to be most effective at 480 nm? Assume that the coating’s index of refraction is less than that of the lens. Express your answer using two significant figures.

1.33

Explanation:

Using the formula for destructive interference since the two reflected ray light waves from eyeglass should be made to cancel each other ( destructive interference) and also refractive index of air less than the refractive index of the film and both are less than the refractive index of the glass meaning the both reflected rays from the air and film will experience a phase change

n film =  ( m + 0.5) ( λ / 2t)

since 2 t = ( m + 0.5) ( λ / n film)

where m is an integer, λ is wavelength, and t is thickness  and n film is the refractive index of film

for effectiveness m = 0

n film = ( 0.5) ( 480 ÷ (2 × 90 ) ) = 1.33

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: